diff options
Diffstat (limited to '3792')
87 files changed, 2044 insertions, 0 deletions
diff --git a/3792/CH1/EX1.1/Ex1_1.sce b/3792/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..feb96ea0b --- /dev/null +++ b/3792/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,31 @@ +// SAMPLE PROBLEM 1/1
+clc;funcprot(0);
+// Given data
+W=100;// lb
+theta=45;// degree
+h=200;// mi
+R=3959;// mi
+g_f=32.1740;// ft/sec^2
+g_m=9.80655;// m/s^2
+g_0=32.234;// ft/sec^2
+m_E=4.095*10^23;// lbf-s^2/ft
+G=3.439*10^-8;// ft^4/(lbf-s^4)
+
+// Calculation
+// (a)
+m_a=W/g_f;// slugs
+W_a=W*4.4482;// N
+m=W_a/g_m;// kg
+printf("\n(a)The mass of the module in slugs,m=%1.2f slugs \n The weight of the module in newtons,W=%3.0f N \n The mass of the module in kilograms,m=%2.1f",m_a,W_a,m);
+// Again using the table inside the front cover, we have
+m=W*0.45359;// kg
+// (b)
+g_h=(g_0*((R^2)/(R+h)^2));
+W_h=m_a*g_h;
+printf("\n(b)The weight at an altitude of 200 miles is then,W_h=%2.1f lb",W_h);
+W_h=W_h*4.4482;
+printf("\n The weight at an altitude of 200 miles is in newton,W_h=%3.0f N",W_h);
+W_h=(G*m_E*m_a)/((R+h)*5280)^2;
+// (c)
+// The weight of an object (the force of gravitational attraction) does not depend on the motion of the object. Thus the answers for part (c) are the same as those in part (b).
+printf("\n(c)The weight of the module in both pounds and newtons,W_h=%2.1f lb (or) %3.0f N",W_h,W_h*4.4482);
diff --git a/3792/CH2/EX2.1/Ex2_1.sce b/3792/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..d95f5300f --- /dev/null +++ b/3792/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,21 @@ +// Example 2_1
+clc;funcprot(0);
+// Given data
+// s=2t^3-24t+6;
+v_a=72;// Velocity in m/s
+v_b=30;// Velocity in m/s
+t_0=1;// s
+t_1=4;// s
+
+// Calculation
+// v=6t^2-24;
+// a=12t;
+// (a)
+t=sqrt((v_a+24)/6);// Time in s
+// (b)
+a=sqrt((v_b+24)/6);// Time in s
+// (c)
+s4=((2*t_1^3)-(24*t)+6);// m
+s1=((2*t_0^3)-(24*t_0)+6)// m;
+deltaS=s4-s1;// The net displacement during the specified interval in m
+printf("\n(a)The time required for the particle to reach a velocity of 72 m/s from its initial condition at t=0 is %1.0f s.\n(b)The acceleration of the particle a=%2.0f m/s^2 \n(c)The net displacement,deltaS=%2.0f m",t,a,deltaS);
diff --git a/3792/CH2/EX2.10/Ex2_10.sce b/3792/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..c52f32337 --- /dev/null +++ b/3792/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,18 @@ +// Example 2_10
+clc;funcprot(0);
+// Given data
+theta_i=30;// degrees
+r=25*10^4;// ft
+rdot=4000;// ft/sec
+theta=0.80;// deg/sec
+g=31.4;// ft/sec^2
+
+// Calculation
+v_r=rdot;// ft/sec
+v_theta=r*(theta*%pi/180);// ft/sec
+v=sqrt(v_r^2+v_theta^2);// ft/sec
+a_r=-g*cosd(theta_i);// ft/sec^2
+a_theta=g*sind(theta_i);// ft/sec^2
+rdotdot=a_r+(r*(theta*(%pi/180))^2);// ft/sec^2
+thetadotdot=(a_theta-(2*rdot*theta*%pi/180))/r;// ft/sec^2
+printf("\nThe velocity of the rocket,v=%4.0f ft/sec \nrdotdot=%2.1f ft/sec^2 and thetadotdot=%1.2e rad/sec^2",v,rdotdot,thetadotdot);
diff --git a/3792/CH2/EX2.12/Ex2_12.sce b/3792/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..3e360eda7 --- /dev/null +++ b/3792/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,32 @@ +// Example 2_12
+clc;funcprot(0);
+// Given data
+v_0=250;// km/h
+theta_i=15;// degree
+a=0.8;// m/s^2
+t=60;// seconds
+s_0=0;// m
+x=3000;// m
+
+// Calculation
+// (a)
+v_0=v_0/3.6;// m/s
+v=v_0+(a*t);// m/s
+s=s_0+(v_0*t)+((1/2)*a*t^2);// m
+y=s*cosd(theta_i);// m
+theta=atand(y/x);// degree
+r=sqrt(x.^2+y.^2);// m
+v_xy=v*cosd(theta_i);// m/s
+v_r=v_xy*sind(theta);// m/s
+v_theta=v_xy*cosd(theta);// m/s
+thetadot=v_theta/r;// rad/s
+zdot=v*sind(theta_i);// m/s
+v_z=zdot;// m/s
+// (b)
+z=y*tand(theta_i);// m
+phi=atand(z/r);// degree
+R=sqrt(r^2+z^2);// m
+v_R=(v_r*cosd(phi))+(zdot*sind(phi));// m/s
+v_phi=(zdot*(cosd(phi)))-(v_r*sind(phi));// m/s
+phidot=v_phi/R;// m/s
+printf("\n(a)v_r=%2.1f m/s \n thetadot=%1.2e rad/s \n zdot=v_z=%2.1f m/s \n(b)v_R=%3.1f m/s \n thetadot=%1.2e rad/s \n phidot=%1.3e rad/s",v_r,thetadot,zdot,v_R,thetadot,phidot);
diff --git a/3792/CH2/EX2.13/Ex2_13.sce b/3792/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..f5de18d19 --- /dev/null +++ b/3792/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,28 @@ +// Example 2_13
+clc;funcprot(0);
+// Given data
+v_A=800;// km/h
+theta_1=45;// degree
+theta_2=60;// degree
+theta_3=75;// degree
+
+// Calculation
+// (I) Graphical.
+v_BA=586;// km/h
+v_B=717;// km/h
+printf("\nv_BA=%3.0f km/h and v_B=%3.0f km/h",v_BA,v_B);
+// (II) Trigonometric.
+v_B=(sind(theta_2)*v_A)/sind(theta_3);// km/h
+printf("\nv_B=%3.0f km/h",v_B);
+// (III) Vector Algebra
+v_B=[(v_B*cosd(theta_1)),(v_B*sind(theta_1))];// km/h
+v_BA=[-(v_BA*cosd(theta_2)),(v_BA*sind(theta_2))];// km/h
+function[X]=velocity(y)
+ X(1)=(v_A-(y(2)*cosd(theta_2)))-(y(1)*cosd(theta_1));
+ X(2)=(y(2)*sind(theta_2))-(y(1)*sind(theta_1));
+endfunction
+y=[100,100];
+z=fsolve(y,velocity);
+v_BA=z(1);// km/h
+v_B=z(2);// km/h
+printf("\nv_AB=%3.0f km/h and v_B=%3.0f km/h",v_BA,v_B);
diff --git a/3792/CH2/EX2.14/Ex2_14.sce b/3792/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..2aefb6540 --- /dev/null +++ b/3792/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,24 @@ +// Example 2_14
+clc;funcprot(0);
+// Given data
+v_A=45;// mi/hr
+v_B=30;// mi/hr
+a_A=3;// ft/sec^2
+theta_1=30;// degree
+theta_2=60;// degree
+rho=440;// The radius of curvature in ft
+
+// Calculation
+// Velocity
+v_A=v_A*(5280/3600);// ft/sec
+v_B=v_B*(5280/3600);// ft/sec
+// By the application of the law of cosines and the law of sines gives
+v_BA=sqrt(v_A^2+v_B^2-(2*v_A*v_B*cosd(theta_2)));// ft/sec
+theta=asind((v_B*sind(theta_2))/v_BA);// degree
+// Acceleration
+a_B=(v_B)^2/rho;// ft/sec^2
+a_BAx=a_B*cosd(theta_1)-a_A;// ft/sec^2
+a_BAy=a_B*sind(theta_1);// ft/sec^2
+a_BA=sqrt(a_BAx^2+a_BAy^2);// ft/sec^2
+beta=asind((a_B*sind(theta_1))/a_BA);// degree
+printf("\nv_BA=%2.1f ft/sec \ntheta=%2.1f degree \na_AB=%1.2f ft/sec^2 \nbeta=%2.1f degree",v_BA,theta,a_BA,beta);
diff --git a/3792/CH2/EX2.15/Ex2_15.sce b/3792/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..43e3ca538 --- /dev/null +++ b/3792/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,13 @@ +// Example 2_15
+clc;funcprot(0);
+// Given data
+v_A=0.3;// m/s
+
+// Calculation
+// Solution (I).
+// v_A=y_A,v_B=y_B
+v_B=-(2*v_A)/3;// m/s
+printf("\nThe velocity of B,v_B=%0.1f m/s",v_B);
+// Solution (II).
+v_B=abs((2/3)*v_A);// m/s
+printf("\nThe velocity of B,v_B=%0.1f m/s (upward)",v_B);
diff --git a/3792/CH2/EX2.2/Ex2_2.sce b/3792/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..70aee775b --- /dev/null +++ b/3792/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,19 @@ +// Example 2_2
+clc;funcprot(0);
+// Given data
+v_x=50;// The initial velocity in ft/sec
+a_x=-10;// The acceleration in ft/sec^2
+t_0=8;// s
+t_1=12;// s
+
+// Calculation
+// v_x=90-10t; ft/sec
+v_x0=(90-(10*t_0));// The velocity in ft/sec
+v_x1=(90-(10*t_1));// The velocity in ft/sec
+// x=-5t^2+90t-80; ft
+x_0=(-5*t_0^2)+(90*t_0)-80;// ft
+x_1=(-5*t_1^2)+(90*t_1)-80;// ft
+// The maximum positive x-coordinate is,then, the value of x for t=9 sec which is
+t=9;// sec
+x_max=(-5*t^2)+(90*t)-80;// ft
+printf("\nThe velocity of the particle for the conditions of t=8 sec and t=12 sec,v_x=%2.0f ft/sec & v_x=%2.0f ft/sec \nThe x-coordinate of the particle for the conditions of t=8 sec and t=12 sec, x=%3.0f ft & x=%3.0f ft \nThe maximum positive x-coordinate reached by the particle,x_max=%3.0f ft",v_x0,v_x1,x_0,x_1,x_max)
diff --git a/3792/CH2/EX2.5/Ex2_5.sce b/3792/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..b8c3b5772 --- /dev/null +++ b/3792/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,31 @@ +// Example 2_5
+clc;funcprot(0);
+// Given data
+// v_x=50-60t;
+// y=100-4t^2;
+// where v_x is in meters per second, y is in meters, and t is in seconds.
+
+// Calculation
+// x=50t-8t^2;
+a_x=-16;// The x-component of the acceleration in m/s^2
+// v_y=-8t; The y-component of the velocity in m/s
+a_y=-8;// The y-component of the acceleration in m/s^2
+// When y=0,
+t=sqrt(100/4);
+v_x=50-(16*t);
+v_y=-8*(t);
+v=sqrt((v_x.^2)+(v_y.^2));// m/s
+a=sqrt(a_x.^2+a_y.^2);// m/s^2
+printf("\nThe velocity,v=%2.0fi+(%2.0fj) m/s \nThe acceleration,a=%2.0fi+(%1.0fj) m/s^2",v_x,v_y,a_x,a_y);
+y=[0,20,40,60,80,100];// m
+for(i=1:6)
+ t(i)=sqrt((100-y(i))/4);// s
+ x(i)=((50*t(i))-(8*t(i).^2));// m
+ v_x(i)=((50*t(i)-(8*t(i).^2)));// m/s
+ v_y(i)=(-8*t(i));// m/s
+ v=sqrt((v_x.^2)+(v_y.^2));// m/s
+ a=sqrt(a_x.^2+a_y.^2);// m/s^2
+end
+plot(x',y,'-.*');
+xlabel('x,m');
+ylabel('y,m');
diff --git a/3792/CH2/EX2.5/Fig2_5.jpg b/3792/CH2/EX2.5/Fig2_5.jpg Binary files differnew file mode 100644 index 000000000..2e362780d --- /dev/null +++ b/3792/CH2/EX2.5/Fig2_5.jpg diff --git a/3792/CH2/EX2.6/Ex2_6.sce b/3792/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..c60b35cdf --- /dev/null +++ b/3792/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,44 @@ +// Example 2_6
+clc;funcprot(0);
+// Given data
+v_0=80;// The launch speed in ft/sec
+theta=35;// The launch angle in degree
+m=8;// lb
+g=32.2;// The acceleration due to gravity in ft/sec^2
+y_0=6;// ft
+x_0=0;// ft
+x=100+30;// ft
+
+// Calculation
+v_x0=v_0*cosd(theta);// ft/sec
+t=(x-x_0)/v_x0;// s
+v_y0=v_0*sind(theta);// ft/sec
+y=(y_0+(v_y0*t))-((1/2)*g*t^2);// ft
+// (a)
+// We now find the flight time by setting
+y_01=20;// ft
+function[X]=time(y)
+ X(1)=((y_0+(v_y0*y(1))-((1/2)*g*y(1)^2)))-y_01;
+endfunction
+y=[10];
+z=fsolve(y,time);
+t_f=z(1);// s
+x=x_0+(v_x0*t_f);// ft
+printf("\n(a)The time duration of the flight,t_f=%1.2f s",t_f);
+//(b)
+printf("\n(b)Thus the point of first impact is (x,y)=(%3.0f,%2.0f)ft",x,y_01);
+// (c)
+v_y=0;// ft
+h=((v_y0^2-v_y^2)/(2*g))+6;// ft
+printf("\n(c)The maximum height above the horizontal field attained by the ball,h=%2.1f ft",h);
+// (d)
+v_x=v_x0;// ft/sec
+v_y=v_y0-(g*t_f);// ft/sec
+printf("\n(d)The impact velocity,v=%2.1f i+(%2.1f j) ft/sec",v_x,v_y);
+x=100+30;// ft (given)
+v_0=75;// ft/sec (given)
+v_x0=v_0*cosd(theta);// ft/sec
+t=(x-x_0)/v_x0;// s
+v_y0=v_0*sind(theta);// ft/sec
+y=(y_0+(v_y0*t))-((1/2)*g*t^2);// ft
+printf("\n The point of impact is (x,y)=(%3.0f,%2.1f)ft",x,y);
diff --git a/3792/CH2/EX2.7/Ex2_7.sce b/3792/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..d54a73078 --- /dev/null +++ b/3792/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,24 @@ +// Example 2_7
+clc;funcprot(0);
+// Given data
+a=3;// m/s^2
+v_A=100;// km/h
+v_C=50;// km/h
+s=120;// m
+
+// Calculation
+v_A=v_A*(1000/3600);// The velocity in m/s
+v_C=v_C*(1000/3600);// The velocity in m/s
+a_t=(1/(2*s))*(v_C.^2-v_A.^2);// The acceleration in m/s^2
+// (a) Condition at A.
+a_n=sqrt(a.^2-(a_t).^2);// The acceleration in m/s^2
+rho_A=v_A.^2/a_n;// The radius of curvature at A in m
+// (b) Condition at B.
+a_n=0;// m/s^2
+a_b=a_n+a_t;// The acceleration at the inflection point B in m/s^2
+// (c) Condition at C.
+rho=150;// The radius of curvature of the hump at C in m
+a_n=v_C.^2/rho;// The normal acceleration in m/s^2
+a=sqrt(a_n.^2+a_t.^2);// The total acceleration at C in m/s^2
+printf("\n(a)The radius of curvature at A,rho=%3.0f m \n(b)The acceleration at the inflection point B,a=%1.2f m/s^2 \n(c)The total acceleration at C,a=%1.2f m/s^2",rho_A,a_b,a)
+
diff --git a/3792/CH2/EX2.8/Ex2_8.sce b/3792/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..8ebf9862f --- /dev/null +++ b/3792/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,22 @@ +// Example 2_8
+clc;funcprot(0);
+// Given data
+g=30;// The acceleration due to gravity in ft/sec^2
+theta=15;// The direction of its trajectory in degree
+v=12000;// The velocity in mi/hr
+a_x=20;// The horizontal component of acceleration in ft/sec^2
+a_y=g;// The downward acceleration component in ft/sec^2
+
+// Calculation
+a_n=(a_y*cosd(theta))-(a_x*sind(theta));// The normal component of acceleration in ft/sec^2
+a_t=(a_y*sind(theta))+(a_x*cosd(theta));// The tangential component of acceleration in ft/sec^2
+// (a)
+v=v*44/30;// ft/sec
+rho=v^2/a_n;// The radius of curvature in ft
+// (b)
+vdot=a_t;// The t-component of acceleration in ft/sec^2
+// (c)
+betadot=v/rho;// The angular rate of line GC in rad/sec
+// (d)
+a=[a_n,a_t];// The total acceleration in ft/sec^2
+printf("\n(a) The radius of curvature,rho=%2.2e ft \n(b)The t-component of acceleration,v_dot=%2.1f ft/sec^2 \n(c)The angular rate of line GC,betadot=%2.2e rad/sec \n(d)The total acceleration,a=%2.1f e_n+%2.1f e_t ft/sec^2",rho,vdot,betadot,a(1),a(2));
diff --git a/3792/CH2/EX2.9/Ex2_9.sce b/3792/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..3c2bc9da8 --- /dev/null +++ b/3792/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,21 @@ +// Example 2_9
+clc;funcprot(0);
+// Given data
+// theta=0.2t+0.02t^3;
+// r=0.2+0.04t^2;
+t=3;// s
+
+// Calculation
+r_3=0.2+(0.04*t^2);// m
+rdot_3=0.08*t;// m/s
+rdotdot_3=0.08;// m/s^2
+theta_3=(0.2*t)+(0.02*t^3);// rad
+thetadot_3=0.2+(0.06*t^2);// rad/s
+thetadotdot_3=0.12*t;// rad/s^2
+v_r=rdot_3;// m/s
+v_theta=r_3*thetadot_3;// m/s
+v=sqrt(v_r^2+v_theta^2);// m/s
+a_r=rdotdot_3-(r_3*thetadot_3^2);// m/s^2
+a_theta=((r_3*thetadotdot_3)+(2*rdot_3*thetadot_3));// m/s^2
+a=sqrt(a_r^2+a_theta^2);// m/s^2
+printf("\nThe magnitudes of the velocity and acceleration of the slider, v=%0.3f m/s and a=%0.3f m/s^2",v,a);
diff --git a/3792/CH3/EX3.1/Ex3_1.sce b/3792/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..737c32d7f --- /dev/null +++ b/3792/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,17 @@ +// SAMPLE PROBLEM 3/1
+clc;clear;funcprot(0);
+// Given data
+m=75;// kg
+T=8300;// The tension in the hoisting cable in N
+g=9.81;// The acceleration due to gravity in m/s^2
+m_ems=750;// The total mass of the elevator, man and scale in kg
+t_0=0;// s
+t_1=3;// s
+
+// Calcaulation
+// SigmaF_y=m*a_y;
+a_y=(T-(m_ems*g))/m_ems;// m/s^2
+// SigmaF_y=m*a_y;
+R=((m*a_y)+(m*g));// N
+v=(1.257*t_1)-(1.257*t_0);// m/s
+printf("\nThe equal and opposite reaction,R=%3.0f N \nThe upward velocity of the elevator,v=%1.2f m/s",R,v);
diff --git a/3792/CH3/EX3.11/Ex3_11.sce b/3792/CH3/EX3.11/Ex3_11.sce new file mode 100644 index 000000000..057d33c53 --- /dev/null +++ b/3792/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,15 @@ +// SAMPLE PROBLEM 3/11
+clc;clear;funcprot(0);
+// Given data
+m=50;// kg
+v_1=4;// m/s
+mu_k=0.30;// The coefficient of kinetic friction
+g=9.81;// The acceleration due to gravity in m/sec^2
+s=10;// m
+theta=15;// degree
+R=474;// N
+
+// Calculation
+U_12=((m*g)*s*sind(theta))-(mu_k*R*(s));// The total work done on the crate during the motion in J
+v_2=sqrt((((1/2)*m*v_1^2)+U_12)/((1/2)*m));// The velocity of the crate in m/s
+printf("\nThe velocity of the crate,v_2=%1.2f m/s",v_2);
diff --git a/3792/CH3/EX3.12/Ex3_12.sce b/3792/CH3/EX3.12/Ex3_12.sce new file mode 100644 index 000000000..e99b56895 --- /dev/null +++ b/3792/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 3/12
+clc;clear;funcprot(0);
+// Given data
+m=80;// kg
+v=72;// km/h
+s=75;// m
+g=9.81;// The acceleration due to gravity in m/sec^2
+mu_sa=0.30;// The coefficient of static friction
+mu_ka=0.28;// The coefficient of kinetic friction
+mu_sb=0.25;// The coefficient of static friction
+mu_kb=0.20;// The coefficient of kinetic friction
+
+// Calculation
+// (a)
+a_1=(v/3.6)^2/(2*s);// m/s^2
+F=m*a_1;// The friction force on the block in N
+U_12=F*s;// The work done in J
+printf("\n(a)The work done by the friction force acting on the crate,U_12=%5.0f J (or) %2.0f kJ",U_12,U_12/1000);
+// (b)
+F_1=mu_sb*m*g;// N
+F_2=mu_kb*m*g;// N
+F=F_2;// N
+a=F/m;// The acceleration in m/s^2
+s=(a/a_1)*s;// The displacement of a crate in m
+U_12=F*s;// The work done in J
+printf("\n(b)The work done by the friction force acting on the crate,U_12=%4.0f J (or) %1.2f kJ",U_12,U_12/1000);
diff --git a/3792/CH3/EX3.13/Ex3_13.sce b/3792/CH3/EX3.13/Ex3_13.sce new file mode 100644 index 000000000..b115ee8e9 --- /dev/null +++ b/3792/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,18 @@ +// SAMPLE PROBLEM 3/13
+clc;clear;funcprot(0);
+// Given data
+m=50;// The mass of the block in kg
+F=300;// N
+x_1=0.233;// m
+k=80;// The spring stifness in N/m
+x=1.2;// m
+y=0.9;// m
+
+// Calculation
+x_2=x_1+x;// m
+U_12=(1/2)*k*(x_1^2-x_2^2);// The work done by the spring force acting on the block in J
+s=sqrt(x^2+y^2)-y;// m
+W=F*s;// The work done in J
+T_1=0;// J
+v=sqrt(((U_12+W)*2)/m);// m/s
+printf("\nThe velocity of the block as it reaches position B,v=%1.2f m/s",v);
diff --git a/3792/CH3/EX3.14/Ex3_14.sce b/3792/CH3/EX3.14/Ex3_14.sce new file mode 100644 index 000000000..02efc42ef --- /dev/null +++ b/3792/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,18 @@ +// SAMPLE PROBLEM 3/14
+clc;clear;funcprot(0);
+// Given data
+F=800;// lb
+v=4;/// ft/sec
+P=6;// The power output of the winch in hp
+P_i=8;// hp
+theta=30;// degree
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+N=F*cosd(theta);// lb
+// SigmaF_x=0;
+T=(P*550)/v;// The tension in the cable in N
+mu_k=(T-(F*sind(theta)))/N;// The coefficient of kinetic friction
+T=(P_i*550)/v;// lb
+a=(T-(N*mu_k)-(F*sind(theta)))*(g/F);// The acceleration in ft/sec^2
+printf("\nThe corresponding instantaneous acceleration of the log,a=%2.2f ft/sec^2",a);
diff --git a/3792/CH3/EX3.15/Ex3_15.sce b/3792/CH3/EX3.15/Ex3_15.sce new file mode 100644 index 000000000..ddf8ea777 --- /dev/null +++ b/3792/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,12 @@ +// SAMPLE PROBLEM 3/15
+clc;clear;funcprot(0);
+// Given data
+h_1=500;// km
+v_1=30000;// km/h
+h_2=1200;// km
+R=6371;// km
+g=9.81;// The acceleration due to gravity in m/sec^2
+
+// Calculation
+v_2=sqrt((v_1/3.6)^2+((2*g*(R*10^3)^2)*((10^-3/(R+h_2))-(10^-3/(R+h_1)))));
+printf("\nThe velocity of the satellite as it reaches point B,v_2=%4.0f m/s (or) v_2=%5.0f km/h",v_2,v_2*3.6);
diff --git a/3792/CH3/EX3.16/Ex3_16.sce b/3792/CH3/EX3.16/Ex3_16.sce new file mode 100644 index 000000000..26376deda --- /dev/null +++ b/3792/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,21 @@ +// SAMPLE PROBLEM 3/16
+clc;clear;funcprot(0);
+// Given data
+mg=6;// lb
+k=2;// lb/in
+g=32.2;// The acceleration due to gravity in ft/sec^2
+h=24;// in
+x_1=24/12;// ft
+x_2=(((24*sqrt(2))/12)-(24/12));// ft
+
+// Calculation
+// The reaction of the rod on the slider is normal to the motion and does no work.
+T_1=0;// ft-lb
+U_12=0;// ft-lb
+// We define the datum to be at the level of position 1, so that the gravitational potential energies are
+V_1g=0;// ft-lb
+V_2g=-(mg)*(h/12);// ft-lb
+V_1e=(1/2)*(k*12)*(x_1)^2;// ft-lb
+V_2e=(1/2)*(k*12)*(x_2)^2;// ft-lb
+v_2=sqrt(((T_1+(V_1g+V_1e)+U_12)-(V_2g+V_2e))*(2*(g/mg)));// ft/sec
+printf("\nThe velocity of the slider as it passes position 2,v_2=%2.1f ft/sec",v_2);
diff --git a/3792/CH3/EX3.17/Ex3_17.sce b/3792/CH3/EX3.17/Ex3_17.sce new file mode 100644 index 000000000..12664cd9e --- /dev/null +++ b/3792/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 3/17
+clc;clear;funcprot(0);
+// Given data
+m=10;// kg
+k=60;// N/m
+F=250;// N
+theta=30;// degree
+ABbar=1.5;// m
+BCbar=0.9;// m
+g=9.81;// The acceleration due to gravity in m/sec^2
+d_AC=1.2;// The distance in m
+d_BC=0.9;// The distance in m
+
+// Calculation
+s=ABbar-BCbar;// m
+U_ac=F*s;// J
+V_Ag=0;// The initial gravitational potential energy in J
+T_A=(1/2)*m*V_Ag^2;// N.m
+V_Cg=m*g*(d_AC*sind(theta));// The final gravitational potential energy in J
+x_A=s;// m
+x_B=s+d_AC;// m
+V_Ae=(1/2)*k*(x_A)^2;// The initial elastic potential energy in J
+V_Ce=(1/2)*k*(x_B)^2;// The final elastic potential energy in J
+// Substitution into the alternative work-energy equation 3/21a gives
+v_c=sqrt((((T_A+V_Ag+V_Ae+U_ac)-(V_Cg+V_Ce))*2)/m);// m/s
+printf("\nThe velocity of the slider as it passes point C,v_C=%0.3f m/s",v_c);
diff --git a/3792/CH3/EX3.18/Ex3_18.sce b/3792/CH3/EX3.18/Ex3_18.sce new file mode 100644 index 000000000..16b01a9a2 --- /dev/null +++ b/3792/CH3/EX3.18/Ex3_18.sce @@ -0,0 +1,28 @@ +// SAMPLE PROBLEM 3/18
+clc;clear;funcprot(0);
+// Given data
+m_A=2;// kg
+m_B=4;// kg
+L=0.5;// m
+K_theta=13;// N.m/rad
+g=9.81;// The acceleration due to gravity in m/sec^2
+
+// Calculation
+// (a)
+// T_1+V_1+U_12=T_2+V_2
+function[X]=velocity(y)
+ X(1)=(((1/2)*m_A*y(1)^2)+((1/2)*m_B*(y(1)/4)^2)-(m_A*g*L)-(m_B*g*(L*sqrt(2)/4))+((1/2)*K_theta*(%pi/2)^2))-0;
+endfunction
+y=[0.1];
+v_A=fsolve(y,velocity);// m/s
+printf("\nThe speed of particle A,v_A=%0.3f m/s",v_A);
+// (b)
+for(i=1:10)
+ theta=[0,10,20,30,40,50,60,70,80,90];// degree
+ // T_1+V_1+U_12=T_2+V_2
+ v_A(i)=sqrt(((m_A*g*L*(1-cosd(theta(i))))+((m_B*g*(1/2)*[((L*sqrt(2))/2)-((2*(L/2)*sind((90-(theta(i)))/2)))]))-((1/2)*K_theta*(theta(i)*(%pi/180))^2))/(((1/2)*m_A)+((1/2)*m_B*((1/4)*cosd((90-(theta(i)))/2))^2)));
+end
+plot(theta',v_A);
+xlabel('theta,deg');
+ylabel('v_A,m/s');
+printf("\nThe maximum value of v_A is seen to be (v_A)_max=1.400 m/s at theta=56.4 degree.");
diff --git a/3792/CH3/EX3.18/Fig3_18.jpg b/3792/CH3/EX3.18/Fig3_18.jpg Binary files differnew file mode 100644 index 000000000..50629ec80 --- /dev/null +++ b/3792/CH3/EX3.18/Fig3_18.jpg diff --git a/3792/CH3/EX3.19/Ex3_19.sce b/3792/CH3/EX3.19/Ex3_19.sce new file mode 100644 index 000000000..202752863 --- /dev/null +++ b/3792/CH3/EX3.19/Ex3_19.sce @@ -0,0 +1,20 @@ +// SAMPLE PROBLEM 3/19
+clc;clear;funcprot(0);
+// Given data
+v_1=50;// ft/sec
+v_2=70;// ft/sec
+theta=15;// degree
+dt=0.02;// sec
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+W=2/16;// N
+v_1x=v_1;// ft/sec
+v_2x=v_2;// ft/sec
+v_1y=0;// ft/sec
+v_2y=v_2;// ft/sec
+R_x=(((W/g)*(v_2x*cosd(theta)))+((W/g)*(v_1x)))/dt;// lb
+R_y=(((W/g)*(v_2y*sind(theta)))+((W/g)*(v_1y)))/dt;// lb
+R=sqrt(R_x^2+R_y^2);// lb
+beta=atand(R_y/R_x);// degree
+printf("\nThe magnitude of the average force exerted by the racket on the ball,R=%2.1f lb \nThe angle made by R with the horizontal,beta=%1.2f degree",R,beta);
diff --git a/3792/CH3/EX3.2/Ex3_2.sce b/3792/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..8b49c7a7c --- /dev/null +++ b/3792/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,17 @@ +// SAMPLE PROBLEM 3/2
+clc;clear;funcprot(0);
+// Given data
+m=200;// The mass of the small inspection car in kg
+T=2.4;// kN
+x=12;// adjacent side
+y=5;// opposite side
+r=13;// hypotenuse side
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+W=(m*g)/1000;// The weight in N
+// SigmaF_y=0;
+P=(T*(y/r))+(W*(x/r));// The total force exerted by the supporting cable on the wheels in N
+// SigmaF_x=ma_x
+a=((T*10^3*(x/r))-(W*10^3*(y/r)))/m;// The acceleration of the car in m/s^2
+printf("\nThe total force exerted by the supporting cable on the wheels,P=%1.2f kN \nThe acceleration of the car,a=%1.2f m/s^2",P,a);
diff --git a/3792/CH3/EX3.20/Ex3_20.sce b/3792/CH3/EX3.20/Ex3_20.sce new file mode 100644 index 000000000..592de95d0 --- /dev/null +++ b/3792/CH3/EX3.20/Ex3_20.sce @@ -0,0 +1,10 @@ +// SAMPLE PROBLEM 3/20
+clc;clear;funcprot(0);
+// Given data
+// G=(3/2)*(t^2+3)j-((2/3)*(t^3-4))k
+t=2;// sec
+
+// Calculation
+F=[3*(t),2-(2*t^2)];// [j,k] lb
+F_r=sqrt(F(1)^2+F(2)^2);// lb
+printf("\nF=%1.0fj+(%1.0fk)lb \nF=%1.3f lb",F(1),F(2),F_r);
diff --git a/3792/CH3/EX3.21/Ex3_21.sce b/3792/CH3/EX3.21/Ex3_21.sce new file mode 100644 index 000000000..9772ccb04 --- /dev/null +++ b/3792/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,17 @@ +// SAMPLE PROBLEM 3/21
+clc;clear;funcprot(0);
+// Given data
+m=0.5;// kg
+v_1x=10;// m/s
+v_1y=0;// m/s
+t_1=1;// s
+t_2=2;// s
+t_3=3;// s
+
+// Calculation
+v_2x=((m*v_1x)-((4*(t_1))+(2*(t_3-t_1))))/(m);// m/s
+v_2y=((m*v_1y)+((1*(t_2))+(2*(t_3-t_2))))/(m);// m/s
+v_2=[v_2x,v_2y];// m/s
+v_2=norm(v_2);// m/s
+theta_x=180+atand(v_2y/v_2x);// degree
+printf("\nThe velocity of the particle at the end of the 3-s interval,v_2=%2.0f m/s \ntheta_x=%3.1f degree",v_2,theta_x);
diff --git a/3792/CH3/EX3.22/Ex3_22.sce b/3792/CH3/EX3.22/Ex3_22.sce new file mode 100644 index 000000000..e6f92681d --- /dev/null +++ b/3792/CH3/EX3.22/Ex3_22.sce @@ -0,0 +1,17 @@ + // SAMPLE PROBLEM 3/22
+clc;funcprot(0);
+// Given data
+m=150;// kg
+v_1=4;// m/s
+t_0=0;// s
+t_1=4;// s
+P=600;// N
+t_2=8;// s
+theta=30;// degree
+g=9.81;// The acceleration due to gravity in m/sec^2
+
+// Calculation
+deltat=(m*0)+((m*v_1)-((v_1*2*P)/2)+(m*g*sind(theta)))/((2*P)+(m*g*sind(theta)));// s
+t_a=v_1+deltat;// s
+v_2x=((m*-v_1)+((v_1*2*P)/2)+(v_1*2*P)-(m*g*sind(theta)*t_2))/m;// m/s
+printf("\n(a)The time at which the skip reverses its direction,t_a=%1.2f s \n(b)The velocity of the skip,v_2x=%1.2f m/s",t_a,v_2x);
diff --git a/3792/CH3/EX3.23/Ex3_23.sce b/3792/CH3/EX3.23/Ex3_23.sce new file mode 100644 index 000000000..f55cb7bc9 --- /dev/null +++ b/3792/CH3/EX3.23/Ex3_23.sce @@ -0,0 +1,16 @@ +// SAMPLE PROBLEM 3/23
+clc;funcprot(0);
+// Given data
+m_1=0.050;// kg
+m_2=4;// kg
+v_1=600;// m/s
+v_2=12;// m/s
+theta=30;// degree
+
+// Calculation
+v_2=[(m_2*v_2*cosd(theta))/(m_1+m_2),((m_1*v_1)+(m_2*v_2*sind(theta)))/(m_1+m_2)];// m/s
+v_x=v_2(1);// m/s
+v_y=v_2(2);// m/s
+V_2=sqrt((v_x^2+v_y^2));// m/s
+theta=atand((v_y/v_x));// degree
+printf("\nThe velocity of the block and embedded bullet immediately after impact,v_2=%2.2fi+%2.2fj m/s \nThe final velocity and its direction are given by v_2=%2.2f m/s and theta=%2.1f degree",v_x,v_y,V_2,theta);
diff --git a/3792/CH3/EX3.24/Ex3_24.sce b/3792/CH3/EX3.24/Ex3_24.sce new file mode 100644 index 000000000..739052428 --- /dev/null +++ b/3792/CH3/EX3.24/Ex3_24.sce @@ -0,0 +1,23 @@ +// SAMPLE PROBLEM 3/24
+clc;funcprot(0);
+// Given data
+F_z=10;// N
+m=2;// kg
+v_y=5;// m/s
+x=3;// m
+y=6;// m
+z=4;// m
+
+// Calculation
+r=[x,y,z];// m
+mv=[m*0,m*v_y,m*0];// (kg.m/s)
+H_O1=det([r(2),r(3);mv(2),mv(3)]);// N.m/s
+H_O2=-det([r(1),r(3);mv(1),mv(3)]);// N.m/s
+H_O3=det([r(1),r(2);mv(1),mv(2)]);// N.m/s
+H_O=[H_O1,H_O2,H_O3];// m/s
+F=[0,0,F_z];// N
+Hdot_O1=det([r(2),r(3);F(2),F(3)]);// N.m
+Hdot_O2=-det([r(1),r(3);F(1),F(3)]);// N.m
+Hdot_O3=det([r(1),r(2);F(1),F(2)]);// N.m
+Hdot_O=[Hdot_O1,Hdot_O2,Hdot_O3];// N.m
+printf("\nThe angular momentum H_O about point O,H_O=%2.0fi+(%2.0f)j+%2.0fk N.m/s \nThe time derivative,Hdot=%2.0fi+(%2.0f)j+%2.0fk N.m",H_O(1),H_O(2),H_O(3),Hdot_O(1),Hdot_O(2),Hdot_O(3));
diff --git a/3792/CH3/EX3.25/Ex3_25.sce b/3792/CH3/EX3.25/Ex3_25.sce new file mode 100644 index 000000000..2dca932fa --- /dev/null +++ b/3792/CH3/EX3.25/Ex3_25.sce @@ -0,0 +1,10 @@ +// SAMPLE PROBLEM 3/25
+clc;funcprot(0);
+// Given data
+v_A=740;// m/s
+r_A=6000*10^6;// km
+r_B=75*10^6;// km
+
+// Calculation
+v_B=(r_A*v_A)/r_B;// m/s
+printf("\nThe speed of comet at the point B of closest approach to the sun,v_B=%5.0f m/s",v_B);
diff --git a/3792/CH3/EX3.28/Ex3_28.sce b/3792/CH3/EX3.28/Ex3_28.sce new file mode 100644 index 000000000..1434aac9d --- /dev/null +++ b/3792/CH3/EX3.28/Ex3_28.sce @@ -0,0 +1,21 @@ +// SAMPLE PROBLEM 3/28
+clc;funcprot(0);
+// Given data
+m=800;// kg
+g=9.81;// m/s^2
+h=2;// m
+m_p=2400;// kg
+h_1=0.1;// m
+
+// Calculation
+v_r=sqrt(2*g*h);// m/s
+v_ra=sqrt(2*g*h_1);// m/s
+// (a)
+v_pa=(((m*v_r)+0)+(m*v_ra))/m_p;// m/s
+// (b)
+e=(v_pa+v_ra)/(v_r+0);// The coefficient of restitution
+// (c)
+T=m*g*h;// J
+T_a=((m*v_ra**2)/2)+((m_p*v_pa**2)/2);// J
+E_l=((T-T_a)/T)*100;// The percentage loss of energy(%)
+printf("\n(a)The velocity of the pile immediately after impact,v_p=%1.2f m/s \n(b)The coefficient of restitution,e=%0.3f \n(c)The percentage loss of energy due to the impact is %2.1f percentage.",v_pa,e,E_l);
diff --git a/3792/CH3/EX3.29/Ex3_29.sce b/3792/CH3/EX3.29/Ex3_29.sce new file mode 100644 index 000000000..4cc4d840e --- /dev/null +++ b/3792/CH3/EX3.29/Ex3_29.sce @@ -0,0 +1,15 @@ +// SAMPLE PROBLEM 3/29
+clc;funcprot(0);
+// Given data
+v_1=50;// m/s
+v_2=0;// m/s
+e=0.5;// The effective coefficient of restitution
+theta=30;// degree
+
+// Calculation
+v_1an=e*v_1*sind(theta);// ft/sec
+v_1at=v_1*cosd(theta);// ft/sec
+// Assume ' as a
+v_a=sqrt((v_1an)**2+(v_1at)**2);// ft/sec
+theta_a=atand((v_1an/v_1at));// degree
+printf("\nThe rebound velocity and its angle are then v_a=%2.1f ft/sec and theta_a=%2.1f degree",v_a,theta_a);
diff --git a/3792/CH3/EX3.3/Ex3_3.sce b/3792/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..793f50e52 --- /dev/null +++ b/3792/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 3/3
+clc;clear;funcprot(0);
+// Given data
+m_A=250;// The mass of concrete block A in lb
+m=400;// lb
+theta=30;// degree
+mu_k=0.5;// The coefficient of kinetic friction between the log and the ramp
+x=20;// ft
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+// SigmaF_y=0;
+N=m*cosd(theta);// lb
+// SigmaF_x=ma_x;
+function[X]=acceleration(y)
+ X(1)=0-((2*y(2))+y(3));
+ X(2)=((mu_k*N)-(2*y(1))+(m*sind(theta)))-((m/g)*y(2));
+ X(3)=(m_A-y(1))-((m_A/g)*y(3));
+endfunction
+y=[100,1,1];
+z=fsolve(y,acceleration);
+T=z(1);// lb
+a_A=z(3);// ft/sec^2
+a_C=z(2);// ft/sec^2
+v_A=sqrt(2*a_A*x);// ft/sec
+printf("\nThe velocity of the block as it hits the ground at B,v_A=%2.2f ft/sec",v_A);
diff --git a/3792/CH3/EX3.30/Ex3_30.sce b/3792/CH3/EX3.30/Ex3_30.sce new file mode 100644 index 000000000..67df7f584 --- /dev/null +++ b/3792/CH3/EX3.30/Ex3_30.sce @@ -0,0 +1,32 @@ +// SAMPLE PROBLEM 3/30
+clc;funcprot(0);
+// Given data
+v_1=6;// m/s
+v_2=0;// m/s
+e=0.6;// The coefficient-of-restitution
+theta=30;// degree
+
+// Calculation
+// Assume a for '
+v_1n=v_1*cosd(theta);// m/s
+v_1t=v_1*sind(theta);// m/s
+v_2n=0;// m/s
+v_2t=v_2n;// m/s
+function[X]=velocity(y)
+ X(1)=(v_1n+v_2n)-(y(1)+y(2));
+ X(2)=(e*(v_1n+v_2n))-(y(2)-y(1));
+endfunction
+y=[1,1];
+z=fsolve(y,velocity);
+v_1an=z(1);// m/s
+v_2an=z(2);// m/s
+v_1at=v_1t;// m/s
+v_2at=v_2t;// m/s
+v_1a=sqrt((v_1an)^2+(v_1at)^2);// m/s
+v_2a=sqrt((v_2an)^2+(v_2at)^2);// m/s
+thetaa=atand(v_1an/v_1at);// m/s
+// The kinetic energies just before and just after impact, with m=m1=m2,are
+T=18;// m
+T_a=13.68;// m
+E_l=((T-T_a)/T)*100;// The percentage energy loss(%)
+printf("\nThe final speeds of the particles v_1a=%1.2f m/s ,v_2a=%1.2f m/s \nThe angle which v_1a makes with the t-direction,theta=%2.2f degree \nThe percentage energy loss is %2.0f percentage.",v_1a,v_2a,thetaa,E_l);
diff --git a/3792/CH3/EX3.31/Ex3_31.sce b/3792/CH3/EX3.31/Ex3_31.sce new file mode 100644 index 000000000..53efcea11 --- /dev/null +++ b/3792/CH3/EX3.31/Ex3_31.sce @@ -0,0 +1,23 @@ +// SAMPLE PROBLEM 3/31
+clc;funcprot(0);
+// Given data
+h_1=2000;// The perigee altitude in km
+h_2=4000;// The apogee altitude in km
+h_c=2500;//The altitude of the satellite in km
+g=9.825;// The acceleration due to gravity in m/sec^2
+R=12742/2;// km
+
+// Calculation
+// (a)
+r_max=R+h_2;// km
+r_min=R+h_1;// km
+a=(r_min+r_max)/2;// km
+v_P=(R*10^3*sqrt(g/(a*10^3))*sqrt(r_max/r_min));// m/s
+v_A=(R*10^3*sqrt(g/(a*10^3))*sqrt(r_min/r_max));// m/s
+// (b)
+r=R+h_c;// km
+v_C=sqrt((2*g*(R*10^3)^2)*((1/r)-(1/(2*a)))*(1/10^3));// m/s
+// (c)
+tau=(2*%pi*((a*10^3)^(3/2)))/((R*10^3)*sqrt(g));// km
+tau_h=tau/3600;// km
+printf("\n(a)The necessary perigee velocity,v_P=%4.0f m/s (or) %5.0f km/h \n The necessary apogee velocity,v_A=%4.0f m/s (or) %5.0f km/h \n(b)The velocity at point C,v_C=%4.0f m/s (or) %5.0f km/h \n(c)The period of the orbit,tau=%1.3f h",v_P,v_P*3.6,v_A,v_A*3.6,v_C,v_C*3.6,tau_h);
diff --git a/3792/CH3/EX3.4/Ex3_4.sce b/3792/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..522d11a72 --- /dev/null +++ b/3792/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,17 @@ +// SAMPLE PROBLEM 3/4
+clc;clear;funcprot(0);
+// Given data
+m=10;// The mass in kg
+v=2;// The speed in m/s
+R=8;// N
+
+// Calculation
+k=R/v^2;// N.s^2/m^2
+// SigmaF_x=ma_x;
+v_0=v;// m/s
+v=v_0/2;// m/s
+t=((1/v)-(1/2));// The time in s
+t_0=0;// s
+t_1=2.5;// s
+x=integrate('10/(5+(2*t))','t',t_0,t_1);
+printf("\nThe corresponding travel distance,x=%1.2f m",x);
diff --git a/3792/CH3/EX3.8/Ex3_8.sce b/3792/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..33d91f508 --- /dev/null +++ b/3792/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,24 @@ +// SAMPLE PROBLEM 3/8
+clc;clear;funcprot(0);
+// Given data
+m=1500;// The mass of the car in kg
+v_A=100;// The velocity in km/h
+v_C=50;// The velocity in km/h
+rho_A=400;// The radius of curvature in m
+rho_C=80;// The radius of curvature in m
+delta_s=200;// m
+
+// Calculation
+a_t=abs((((v_C/3.6)^2)-((v_A/3.6)^2))/(2*delta_s));// The tangential acceleration in m/s^2
+a_na=((v_A/3.6)^2)/rho_A;// The normal components of acceleration at A in m/s^2
+a_nb=0;// The normal components of acceleration at B in m/s^2
+a_nc=((v_C/3.6)^2)/rho_C;// The normal components of acceleration at C in m/s^2
+F_t=m*a_t;// N
+F_na=m*a_na;// N
+F_nb=m*a_nb;// N
+F_nc=m*a_nc;// N
+F_a=sqrt(F_na^2+F_t^2);// The total horizontal force acting on the tires at A in N
+F_b=sqrt(F_nb^2+F_t^2);// The total horizontal force acting on the tires at B in N
+F_c=sqrt(F_nc^2+F_t^2);// The total horizontal force acting on the tires at C in N
+printf("\nAt A,F=%4.0f N \nAt B,F=%4.0f N \nAt C,F=%4.0f N",F_a,F_b,F_c);
+
diff --git a/3792/CH3/EX3.9/Ex3_9.sce b/3792/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..38136a4c9 --- /dev/null +++ b/3792/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,11 @@ +// SAMPLE PROBLEM 3/9
+clc;clear;funcprot(0);
+// Given data
+h=200;// The altitude in mi
+R=3959;// mi
+g=32.234;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+// SigmaF_n=ma_n;
+v=(R*5280)*sqrt(g/((R+h)*5280));// ft/sec
+printf("\nThe velocity required for the spacecraft,v=%5.0f ft/sec",v);
diff --git a/3792/CH4/EX4.4/Ex4_4.sce b/3792/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..409498d73 --- /dev/null +++ b/3792/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,24 @@ +// SAMPLE PROBLEM 4/4
+clc;clear;funcprot(0);
+// Given data
+m=20;// kg
+u_z=300;// m/s
+g=9.81;// m/s^2
+m_a=5;// kg
+m_b=9;// kg
+m_c=6;// kg
+theta=45;// degree
+s=4000;// m
+x=3;// m
+y=4;// m
+r=5;// m
+h_a=500;// m
+
+// Calculation
+t=(u_z*(y/r))/g;// The time required for the shell to reach P in s
+h=u_z^2/(2*g);// The verticl rise in m
+v_a=sqrt(2*g*h_a);// m/s
+v_b=s/t;// m/s
+v_c=[(m*u_z*(x/r))-(m_b*v_b*cosd(theta)),(m_b*v_b*sind(theta)),(m_a*v_a)]/6;// m/s
+v_c=sqrt((v_c(1))^2+(v_c(2))^2+(v_c(3))^2);// m/s
+printf("\nThe velocity which fragment C has immediately after the explosion,v_C=%3.0f m/s",v_c);
diff --git a/3792/CH4/EX4.5/Ex4_5.sce b/3792/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..ec601cd5d --- /dev/null +++ b/3792/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,25 @@ +// SAMPLE PROBLEM 4/5
+clc;clear;funcprot(0);
+// Given data
+g=32.2;// The acceleration due to gravity in ft/sec^2
+n_12=80;// rev/min
+n_34=100;// rev/min
+W_a=32.2;// lb
+W_b=3.22;// lb
+n=4;// Number of balls
+vbar=4;// m/s
+r_12=18/12;// ft
+r_34=12/12;// ft
+
+// Calculation
+// (a)Kinetic energy
+v_rel12=r_12*((2*%pi*n_12)/60);// ft/sec
+v_rel34=r_34*((2*%pi*n_34)/60);// ft/sec
+ke=(1/2)*((W_a/g)+(n*(W_b/g)))*(vbar)^2;// ft-lb
+ke_r=(2*[(1/2)*(W_b/g)*v_rel12^2])+(2*[(1/2)*(W_b/g)*v_rel34^2]);// The rotational part of the kinetic energy in ft-lb
+T=ke+ke_r;// The total kinetic energy in ft-lb
+// (b)Linear momentum
+G=((W_a/g)+(n*(W_b/g)))*vbar;// ft-lb-sec
+// (c)Angular momentum about O.
+H_O=(2*[(W_b/g)*r_12*v_rel12])-(2*[(W_b/g)*r_34*v_rel34]);// lb-sec
+printf("\n(a)The kinetic energy,T=%2.0f ft-lb \n(b)The magnitude of the linear momentum,G=%1.1f lb-sec \n(c)The magnitude of the angular momentum about point O,H_O=%1.3f ft-lb-sec",T,G,H_O);
diff --git a/3792/CH5/EX5.1/Ex5_1.sce b/3792/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..592e9c8d4 --- /dev/null +++ b/3792/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,28 @@ +// SAMPLE PROBLEM 5/1
+clc;clear;funcprot(0);
+// Given data
+n_1=1800;// rev/min
+t_0=0;// s
+// alpha=4t;
+n_2=900;// rev/min
+
+// Calculation
+// (a)
+omega_1=(-2*%pi*n_1)/60;// rad/s
+// omega=-(60*%pi)+2t^2
+omega_2=(-2*%pi*n_2)/60;// rad/s
+t=sqrt((omega_2-omega_1)/2);// s
+// (b)
+// The flywheel changes direction when its angular velocity is momentarily zero. Thus,
+t_b=sqrt((0-omega_1)/2);// s
+// (c)
+t_0=0;// s
+t_1=t_b;// s
+theta_1=integrate('omega_1+(2*t^2)','t',t_0,t_1);// rad
+N_1=abs(-theta_1/(2*%pi));// rev(clockwise)
+t_1=t_b;// s
+t_2=14;// s
+theta_2=integrate('omega_1+(2*t^2)','t',t_1,t_2);// rad
+N_2=theta_2/(2*%pi);// rev
+N=N_1+N_2;// rev
+printf("\n(a)The time required for the flywheel to reduce its clockwise angular speed,t=%1.2f s \n(b)The time required for the flywheel to reverse its direction of rotation,t=%1.2f s \n(c)The total number of revolutions,N=%3.0f rev",t,t_b,N);
diff --git a/3792/CH5/EX5.10/Ex5_10.sce b/3792/CH5/EX5.10/Ex5_10.sce new file mode 100644 index 000000000..b9b73f25f --- /dev/null +++ b/3792/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,12 @@ +// SAMPLE PROBLEM 5/10
+clc;funcprot(0);
+// Given data
+v_B=0.8;// The velocity in m/s
+theta=30;// degree
+d_co=18;// The distance in inch
+
+// Calculation
+v_A=v_B*cosd(theta);// ft/sec
+OAbar=(d_co/12)/(cosd(theta));// ft
+omega=v_A/(OAbar);// rad/sec CCW
+printf("\nThe angular velocity of the slotted arm,omega=%0.3f rad/sec CCW",omega);
diff --git a/3792/CH5/EX5.11/Ex5_11.sce b/3792/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..f35879c78 --- /dev/null +++ b/3792/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,14 @@ +// SAMPLE PROBLEM 5/11
+clc;funcprot(0);
+// Given data
+r=300/1000;// m
+r_0=200/1000;// m
+v_o=3;// m/s
+OCbar=r;// m
+theta=120;// degree
+
+// Calculation
+omega=v_o/OCbar;// rad/s
+ACbar=sqrt(r^2+r_0^2-(2*r*r_0*cosd(theta)));// m
+v_A=ACbar*omega;// m/s
+printf("\nThe velocity of point A for the position indicated,v_A=%1.2f m/s",v_A);
diff --git a/3792/CH5/EX5.12/Ex5_12.sce b/3792/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..dc646b95f --- /dev/null +++ b/3792/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,15 @@ +// SAMPLE PROBLEM 5/12
+clc;funcprot(0);
+// Given data
+omega_OB=10;// rad/sec
+theta=45;// degree
+OBbar=(6*sqrt(2))/12;// ft
+BCbar=(14*sqrt(2))/12;// ft
+ACbar=14/12;// ft
+CDbar=15.23/12;// ft
+
+// Calculation
+omega_BC=(OBbar*omega_OB)/BCbar;// rad/sec CCW
+v_A=ACbar*omega_BC;// ft/sec
+v_D=CDbar*omega_BC;// ft/sec
+printf("\nThe velocity of A,v_A=%1.2f ft/sec \nThe velocity of D,v_D=%1.2f ft/sec \nThe angular velocity of link AB,omega_AB=%1.2f rad/sec CCW",v_A,v_D,omega_BC);
diff --git a/3792/CH5/EX5.14/Ex5_14.sce b/3792/CH5/EX5.14/Ex5_14.sce new file mode 100644 index 000000000..2940b99fa --- /dev/null +++ b/3792/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,30 @@ +// SAMPLE PROBLEM 5/14
+clc;clear;funcprot(0);
+// Given data
+omega_CB=2;// rad/s
+r_A=100;// mm
+r_B=75;// mm
+OCbar=250;// mm
+
+// Calculation
+omega_AB=-6/7;// rad/s
+omega_OA=-3/7;// rad/s
+// The acceleration equation is a_A=a_B+(a_A/B)_n+(a_A/B)_t;
+// a_A=(alpha_OA*r_A)+(omega_OA*(omega_OA*r_A))
+// a_A=(-100*alpha_OA)i-((100)*(3/7)^2)j mm/s^2
+// a_B=(alpha_CB*r_B)+(omega_CB*(omega_CB*r_B)) mm/s^2
+// a_B=300i mm/s^2
+// (a_A/B)n=omega_AB*(omega_AB*r_AB)
+// (a_A/B)n=(6/7)^2*(175i-50j) mm/s^2
+// (a_A/B)t= alpha_AB*r_A/B
+// (a_A/B)t=(-50*alpha_AB)i-(175*alpha_AB)j mm/s^2
+// Equate separately the coefficients of the i-terms and the coefficients of the j-terms to give
+function[X]=acceleration(y)
+ X(1)=(-100*y(1))-(429-(50*y(2)));
+ X(2)=(-18.37)-(-36.7-(175*y(2)));
+endfunction
+y=[0.1 1];
+z=fsolve(y,acceleration);
+alpha_AB=z(2);// mm/s^2
+alpha_OA=z(1);// mm/s^2
+printf("\nThe angular acceleration of link AB,alpha_AB=%0.4f rad/s^2 \nThe angular acceleration of link OA,alpha_OA=%1.2f rad/s^2",alpha_AB,alpha_OA);
diff --git a/3792/CH5/EX5.15/Ex5_15.sce b/3792/CH5/EX5.15/Ex5_15.sce new file mode 100644 index 000000000..9e7b29fc3 --- /dev/null +++ b/3792/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 5/15
+clc;funcprot(0);
+// Given data
+N=1500;// rev/min
+theta_1=60;// degree
+r=5/12;// ft
+ABbar=14/12;// ft
+
+// Calculation
+omega=(2*%pi*N)/60;// rad/s
+a_B=r*omega^2;// ft/sec^2
+omega_AB=29.5;// rad/sec
+a_AB_n=ABbar*omega_AB^2;
+// If we adopt an algebraic solution using the geometry of the acceleration polygon, we first compute the angle between AB and the horizontal. With the law of sines, this angle becomes 18.02 degree.
+theta_2=18.02;// degree
+function[X]=acceleration(y)
+ X(1)=((a_B*cosd(theta_1))+(a_AB_n*cosd(theta_2))-(y(2)*sind(theta_2)))-y(1);
+ X(2)=((a_B*sind(theta_1))-(a_AB_n*sind(theta_2))-(y(2)*cosd(theta_2)))-0;
+endfunction
+y=[1000 1000];
+z=fsolve(y,acceleration)
+a_AB_t=z(2);// ft/sec^2
+a_A=z(1);// ft/sec^2
+r=ABbar;// ft
+alpha_AB=a_AB_t/r;// rad/sec^2
+printf("\nThe acceleration of the piston A,a_A=%4.0f ft/sec^2 \nThe angular acceleration of the connecting rod AB,alpha_AB=%4.0f rad/sec^2",a_A,alpha_AB);
diff --git a/3792/CH5/EX5.16/Ex5_16.sce b/3792/CH5/EX5.16/Ex5_16.sce new file mode 100644 index 000000000..f25db2972 --- /dev/null +++ b/3792/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 5/16
+clc;funcprot(0);
+// Given data
+omega=4;// rad/sec
+omegadot=10;// rad/sec^2
+r=6;// in
+rdot=5;// in/sec
+rdotdot=81;// in/sec^2
+
+// Calculation
+// Velocity
+v_rel=rdot;// (k) in/sec
+v_A=[v_rel,(omega*r)];// in/sec
+printf("\nv_A=%1.0fi+%2.0fj in/sec",v_A(1),v_A(2));
+v_A=norm(v_A);// in/sec
+printf("\nv_A=%2.1f in/sec",v_A);
+// Acceleration
+// Assume O=omega*(omega*r);O_1=omegadot*r;O_2=(2*omega*v_rel);
+O=-(omega*(omega*r));// in/sec^2
+O_1=-omegadot*r;// in/sec^2
+O_2=2*(omega)*(v_rel);// in/sec^2
+a_rel=rdotdot;// in/sec^2
+a_A=[(a_rel+O),(O_2+O_1)];// in/sec^2
+printf("\na_A=%2.0fi+(%2.0f)j in/sec^2",a_A(1),a_A(2));
+a_A=norm(a_A);// in/sec^2
+printf("\na_A=%2.0f in/sec",a_A);
diff --git a/3792/CH5/EX5.17/Ex5_17.sce b/3792/CH5/EX5.17/Ex5_17.sce new file mode 100644 index 000000000..238d910c4 --- /dev/null +++ b/3792/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 5/17
+clc;clear;funcprot(0);
+// Given data
+omega=2;// rad/sec
+theta=45;// degree
+OCbar=450;// mm
+CAbar=225;// mm
+
+// Calculation
+// v_A=omega_CA*r_CA;
+// v_A=(225/sqrt(2))omega_CA*(i-j)
+OPbar=sqrt((OCbar-CAbar)^2+(CAbar)^2);// mm
+r=OPbar;// mm
+omega=omega;//(k) rad/s
+O=omega*r;// mm/s
+// Substitution into the relative-velocity equation gives
+// (225/sqrt(2))omega_CA*(i-j)=(450*sqrt(2)j+xdoti)
+// Equating separately the coefficients of the i and j terms yields
+omega_CA=O/(225/sqrt(2));// mm/s
+xdot=(225/sqrt(2))*omega_CA;// mm/s
+v_rel=xdot;// mm/s
+v_A=CAbar*abs(omega_CA);// mm/s
+v_P=OPbar*omega;// mm/s
+v_AP=abs(v_rel);// mm/s
+omega_AC=v_A/CAbar;// rad/s
+printf("\nThe actual angular velocity of CA,omega_CA=%1.0f rad/s \nThe velocity of A relative to the rotating slot in OD,xdot=v_rel=%3.2f mm/s \nThe velocity of pin A,v_A=%3.0f mm/s",omega_CA,xdot,v_A);
diff --git a/3792/CH5/EX5.18/Ex5_18.sce b/3792/CH5/EX5.18/Ex5_18.sce new file mode 100644 index 000000000..ed54dad1a --- /dev/null +++ b/3792/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,25 @@ +// SAMPLE PROBLEM 5/18
+clc;clear;funcprot(0);
+// Given data
+omega=2;// rad/s
+theta=45;// degree
+OCbar=450;// mm
+CAbar=225;// mm
+
+// Calculation
+// a_A=(omegadot*r)+(omega*(omega*r))+(2*omega*v_rel)+a_rel
+// a_A=(omegadot_CA*r_CA)+omega_CA*(omega_CA*r_CA)
+// a_A=[omegadot_CA*(225/sqrt(2))*(-i-j)]-[4k*(-4k*225/sqrt(2))*(-i-j)]
+omega=2;// rad/s
+r=CAbar*sqrt(2);// mm
+omega_CA=-4;// rad/s
+v_rel=(-OCbar*sqrt(2));// mm/s
+// Assume O=omega*(omega*r);O_1=omegadot*r;O_2=(2*omega*v_rel);
+O_1=0;// mm/s^2
+O_2=omega*(omega*r);// mm/s^2
+O_2=2*omega*v_rel;// mm/s^2
+// a_rel=xdotdot;
+// [(1/sqrt(2))*(225omegadot_CA+3600)i]+[(1/sqrt(2))*(-225omegadot_CA+3600)j] =(900*sqrt(2))i-(1800*sqrt(2))j+xdotdoti
+omegadot_CA=(((-1800*sqrt(2))*sqrt(2))-3600)/-225;// rad/s^2
+xdotdot=(((225*omegadot_CA)+3600)/sqrt(2))-(-900*sqrt(2));// mm/s^2
+printf("\nThe angular acceleration of AC,omega_CA=%2.0f rad/s \nThe acceleration of A relative to the rotating slot in OD,xdotdot=%4.0f mm/s",omegadot_CA,xdotdot);
diff --git a/3792/CH5/EX5.19/Ex5_19.sce b/3792/CH5/EX5.19/Ex5_19.sce new file mode 100644 index 000000000..246d02896 --- /dev/null +++ b/3792/CH5/EX5.19/Ex5_19.sce @@ -0,0 +1,19 @@ +// SAMPLE PROBLEM 5/19
+clc;clear;funcprot(0);
+// Given data
+v_B=150;// (i) m/s
+v_A=100;// (i) m/s
+rho=400;// m
+r=-100;// m
+
+// Calculation
+omega=v_B/rho;// (k) rad/s
+r_AB=r;// (j) m
+v_rel=[v_A-(v_B+(-(omega*r)))];// (i) m/s
+a_A=0;// m/s^2
+a_B=(v_B(1))^2/rho;// m/s^2
+omegadot=0;// rad/s
+a_rel=a_A-[a_B+(omegadot*r)+(omega*-(omega*r))+(2*(omega*v_rel))];// m/s^2
+printf("\nThe instantaneous velocity,v_rel=%2.1fi m/s \nThe instantaneous acceleration,a=%1.2fk m/s^2",v_rel,a_rel);
+v_AB=v_A-v_B;// (i) m/s
+a_AB=a_A-a_B;// (j) m/s^2
diff --git a/3792/CH5/EX5.2/Ex5_2.sce b/3792/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..eaf08263b --- /dev/null +++ b/3792/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,24 @@ +// SAMPLE PROBLEM 5/2
+clc;clear;funcprot(0);
+// Given data
+v=3;// ft/sec
+s=4;// ft
+d_C=48;// inch
+d_B=36;// inch
+d_A=12;// inch
+r_A=d_A/2;// inch
+r_C=d_C/2;// inch
+r_B=d_B/2;// inch
+
+// Calculation
+// (a)
+a=v^2/(2*s);// ft/sec^2
+a_t=a;// ft/sec^2
+a_n=v^2/(r_C/12);// ft/sec^2
+a_C=sqrt(a_n^2+a_t^2);// ft/sec^2
+// (b)
+omega_B=v/(r_C/12);// rad/sec
+alpha_B=a_t/(r_C/12);// rad/sec^2
+omega_A=(r_B/r_A)*omega_B;// rad/sec CW
+alpha_A=(r_B/r_A)*alpha_B;// rad/sec^2 CW
+printf("\n(a)The acceleration of point C on the cable in contact with the drum,a_C=%1.2f ft/sec^2 \n(b)The angular velocity and angular accelerationof the pinion A,omega_A=%1.1f rad/sec CW and alpha_A=%1.3f rad/sec^2 CW",a_C,omega_A,alpha_A);
diff --git a/3792/CH5/EX5.3/Ex5_3.sce b/3792/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..93e8009bd --- /dev/null +++ b/3792/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,19 @@ +// SAMPLE PROBLEM 5/3
+clc;clear;funcprot(0);
+// Given data
+alpha=4;// rad/s^2
+omega=-2;// rad/s
+x=0.4;// m
+y=0.3;// m
+
+// Calculation
+// Using the right-hand rule gives
+// omega=-2k rad/s and alpha=+4k rad/s^2
+r=[x,y];// m
+v=[-omega*r(2),omega*r(1)];// (i,j) (k*i=j)(k*j=-i) m/s
+a_n=[-omega*v(2),omega*v(1)];// m/s^2
+a_t=[-alpha*r(2),alpha*r(1)];// m/s^2
+a=a_n+a_t;// m/s^2
+printf("\nThe vector expression for the velocity,v=%0.1fi+(%0.1f)j m/s \nThe vector expression for the acceleration of point A,a=%2.1fi+%0.1fj m/s^2",v(1),v(2),a(1),a(2));
+v=norm(v);// m/s
+a=norm(a);// m/s^2
diff --git a/3792/CH5/EX5.5/Ex5_5.sce b/3792/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..38a2221d2 --- /dev/null +++ b/3792/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,42 @@ +// SAMPLE PROBLEM 5/5
+clc;funcprot(0);
+// Given data
+r_1=4;// inch
+r_2=4;// inch
+// Case(a)
+// Pulley 1:
+omega_1a=0;// rad/sec
+omega_dot=0;// rad/sec
+alpha_1a=omega_dot;
+// Pulley 2:
+omega_2a=2;// rad/sec
+alpha_2a=-3;// rad/sec^2
+// Case(b)
+// Pulley 1:
+omega_1b=1;// rad/sec
+alpha_1b=4;// rad/sec^2
+// Pulley 2:
+omega_2b=2;// rad/sec
+alpha_2b=-2;// rad/sec^2
+ABbar=12;// inch
+AObar=4;// inch
+
+// Calculation
+// Case (a)
+v_D=r_2*omega_2a;// in/sec
+a_D=r_2*alpha_2a;// in/sec
+omega=v_D/ABbar;// rad/sec
+alpha=a_D/ABbar;// in/sec^2
+v_O=AObar*omega;// rad/sec (CCW)
+a_O=AObar*alpha;// rad/sec^2 (CW)
+printf("\n(a)omega=%0.3f rad/sec (CCW)\n alpha=%1.0f rad/sec^2 (CW) \n v_O=%1.3f in/sec \n a_O=%1.0f in/sec^2",omega,alpha,v_O,a_O);
+// Case (b)
+v_C=r_1*omega_1b;// in/sec
+v_D=r_2*omega_2b;// in/sec
+a_C=r_1*alpha_1b;// in/sec^2
+a_D=r_2*alpha_2b;// in/sec^2
+omega=(v_D-v_C)/ABbar;// rad/sec (CCW)
+alpha=(a_D-a_C)/ABbar;// rad/sec^2 (CW)
+v_O=v_C+(AObar*omega);// in/sec
+a_O=a_C+(AObar*alpha);// in/sec
+printf("\n(b)omega=%0.3f rad/sec (CCW)\n alpha=%1.0f rad/sec^2 (CW) \n v_O=%1.3f in/sec \n a_O=%1.0f in/sec^2",omega,alpha,v_O,a_O);
diff --git a/3792/CH5/EX5.6/Ex5_6.sce b/3792/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..e04a12631 --- /dev/null +++ b/3792/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,13 @@ +// SAMPLE PROBLEM 5/6
+clc;funcprot(0);
+// Given data
+v_A=0.3;// m/s
+b=0.2;// m
+theta=30;// degree
+
+// Calculation
+v_B=-v_A*tand(theta);// m/s
+a_B=-((v_A^2)/b)*(secd(theta))^3;// m/s^2
+omega=(v_A/b)*secd(theta);// rad/s
+alpha=((v_A^2)/b^2)*(secd(theta))^2*tand(theta);// rad/s^2
+printf("\nThe velocity of the center of the roller B in the horizontal guide,v_B=%1.4f m/s \nThe acceleration of the center of the roller B in the horizontal guide,a_B=%0.3f m/s^2 \nThe angular velocity of edge CB,omega=%1.3f rad/s \nThe angular acceleration of edge CB,alpha=%1.3f rad/sec^2",v_B,a_B,omega,alpha);
diff --git a/3792/CH5/EX5.7/Ex5_7.sce b/3792/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..a5a6012bc --- /dev/null +++ b/3792/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,30 @@ +// SAMPLE PROBLEM 5/7
+clc;funcprot(0);
+// Given data
+r=0.300;// m
+v_O=3;// m/s
+theta=30;// degree
+r_0=0.200;// m
+ACbar=0.436;// m
+OCbar=0.300;// m
+
+// Calculation
+// Solution I (Scalar-Geometric)
+omega=v_O/r;// rad/s
+v_AO=r_0*omega;// m/s
+v_A=sqrt(v_O^2+v_AO^2+(2*v_O*v_AO*cosd(theta)));// m/s
+v_AC=(ACbar/OCbar)*v_O;// m/s
+v_A=v_AC;// m/s
+printf("\nThe velocity of point A on the wheel,v_A=%1.2f m/s",v_A);
+// Solution II (Vector)
+omega=[0,0,-omega];// rad/s
+r_0=[(r_0*-cosd(theta)),(r_0*sind(theta)),0];// m
+v_O=[v_O,0,0];// m/s
+v_AO1=det([omega(2),omega(3);r_0(2),r_0(3)]);// m/s
+v_AO2=-det([omega(1),omega(3);r_0(1),r_0(3)]);// m/s
+v_AO3=det([omega(1),omega(2);r_0(1),r_0(2)]);// m/s
+v_AO=[v_AO1,v_AO2,v_AO3];// m/s
+v_A=v_O+v_AO;// m/s
+printf("\nThe velocity of point A on the wheel,v_A=%1.0fi+%1.3fj m/s",v_A(1),v_A(2));
+v_A=norm(v_A);// m/s
+
diff --git a/3792/CH5/EX5.8/Ex5_8.sce b/3792/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..4bee5ac56 --- /dev/null +++ b/3792/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,33 @@ +// SAMPLE PROBLEM 5/8
+clc;funcprot(0);
+// Given data
+OCbar=0.250;// m
+omega=2;// rad/s
+OAbar=0.100;// m
+OBbar=0.050;// m
+ABbar=0.075;// m
+
+// Calculation
+// Solution I (Vector)
+r_A=[0,0.100,0];// (i,j,k) m
+r_B=[-0.75,0,0];// (i,j,k) m
+r_AB=[-0.175,0.50,0];// (i,j,k) m
+// omega_OA*r=(omega_CB*r_B)+(omega_AB*r_AB);
+// omega_OA=omega_OA*k
+// omega_CB=2k
+// omega_AB=omega_ABk
+// Matching coefficients of the respective i- and j-terms gives
+omega_AB=-(25*6)/(25*7);// rad/s
+omega_OA=(50*omega_AB)/100;// rad/s
+printf("\n(I)The angular velocity of OA,omega_OA=%0.3f rad/s \n The angular velocity of AB,omega_AB=%0.3f rad/s",omega_OA,omega_AB);
+// Solution II (Scalar-Geometric)
+r_A=0.100;// m
+r_B=0.075;// m
+v_B=r_B*omega;// m/s
+tantheta=(OAbar-OBbar)/(OCbar-r_B);
+// v_AB=v_B/ cos(theta);
+// ABbar= (OCbar-r_AB)/ cos(theta);
+v_A=v_B*tantheta;// m/s
+omega_AB=(v_B)/(OCbar-r_B);// rad/s CW
+omega_OA=v_A/OAbar;// rad/s CW
+printf("\n(II)The angular velocity of OA,omega_OA=%0.3f rad/s \n The angular velocity of AB,omega_AB=%0.3f rad/s",omega_OA,omega_AB);
diff --git a/3792/CH5/EX5.9/Ex5_9.sce b/3792/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..032e74b8f --- /dev/null +++ b/3792/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,27 @@ +// SAMPLE PROBLEM 5/9
+clc;funcprot(0);
+// Given data
+n=1500;// rev/min
+theta=60;// degree
+r=5;// inch
+d_AG=10;// The distance from A to G in inch
+d_GB=4;// The distance from G to B in inch
+d_AB=14;// The distance from A to B in inch
+
+// Calculation
+v_B=(r/12)*((2*%pi*n)/60);// ft/sec
+// From the law of sines,
+beta=asind(r/(d_AB/sind(theta)));// degree
+theta_3=30;// degree
+theta_1=90-beta;// degree
+theta_2=180-theta_3-theta_1;// degree
+v_A=(v_B*sind(theta_2))/sind(theta_1);// ft/sec
+v_AB=(v_B*sind(theta_3))/sind(theta_1);// ft/sec
+ABbar=d_AB/12;// ft
+omega_AB=v_AB/ABbar;// rad/sec
+GBbar=d_GB/12;// ft
+v_GB=(GBbar/ABbar)*v_AB;// ft/sec
+// From velocity diagram
+v_G=64.1;// ft/sec
+printf("\nThe velocity of the piston A,v_A=%2.1f ft/sec \nThe velocity of point G on the connecting rod,v_G=%2.1f ft/sec \nThe angular velocity of the connecting rod,omega_AB=%2.1f rad/sec",v_A,v_G,omega_AB);
+
diff --git a/3792/CH6/EX6.1/Ex6_1.sce b/3792/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..a2f14c658 --- /dev/null +++ b/3792/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,37 @@ +// SAMPLE PROBLEM 6/1
+clc;funcprot(0);
+// Given data
+W=3220;// lb
+v=44;// m/s (30 mi/hr)
+s=200;// ft
+mu=0.8;// The effective coefficient of friction between the tires and the road
+g=32.2;// The acceleration due to gravity in ft/sec^2
+d_G=24;// inch
+d_BG=60;// inch
+d_GA=60;// inch
+
+// Calculation
+abar=v^2/(2*s);// ft/sec^2
+theta=atand(1/10);// degree
+W_h=W*cosd(theta);// lb
+W_v=W*sind(theta);// lb
+mabar=(W/g)*abar;// lb
+// SigmaF_x = m*abar_x
+F=mabar+W_v;// lb
+function[X]=reaction(y)
+ X(1)=(y(1)+y(2)-W)-0;
+ X(2)=((d_GA*y(1))+(F*d_G)-(y(2)*d_BG))-0;
+endfunction
+y=[1000,1000];
+z=fsolve(y,reaction);
+N_1=z(1);// lb
+N_2=z(2);// lb
+FbyN_2=F/N_2;
+printf("\nThe friction force under the rear driving wheels,F=%3.0f lb \nThe normal force under each pair of wheels,N_1=%4.0f lb & N_2=%4.0f lb",F,N_1,N_2);
+// Alternative solution
+// SigmaM_A=m*abar*d
+// SigmaM_A=m*abar*d
+N_2=((mabar*d_G)+((d_GA*W_h)+(d_G*W_v)))/(d_BG+d_GA);// lb
+// SigmaM_B=m*abar*d;
+N_1=((W_h*d_BG)-(d_G*W_v)-(mabar*d_G))/(d_BG+d_GA);// lb
+printf("\nALTERNATIVE SOLUTION:The normal force under each pair of wheels,N_1=%4.0f lb & N_2=%4.0f lb",N_1,N_2);
diff --git a/3792/CH6/EX6.10/Ex6_10.sce b/3792/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..98c39433f --- /dev/null +++ b/3792/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,34 @@ +// SAMPLE PROBLEM 6/10
+clc;funcprot(0);
+// Given data
+l=4;// ft
+W=40;// The weight of the slender bar in N
+theta=30;// degree
+k=30;// The stiffness of the spring in lb/in
+ABbar=24;// inch
+BDbar=24;// inch
+h=-2;// inch
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+
+// Calculation
+// (a)
+// T=[[(1/2)*m*v^2]+((1/2)*I_G*omega^2)];
+// T=1.449*omega^2;
+T_1=0;// ft-lb
+U_12=0;// ft-lb
+V_1=0;// ft-lb
+V_2=W*((2*cosd(theta))-2);// ft-lb
+// We now substitute into the energy equation and obtain
+omega=sqrt(((T_1+V_1+U_12)-(V_2))/1.449);// rad/sec
+// (b)
+x=ABbar-18;// ft
+V_1=0;// ft-lb
+V_3=(1/2)*k*(x^2)/12;// ft-lb
+// T=(1/2)*I_A*omega^2;
+// T_3=0.828*v_B^2;
+U_13=0;// ft-lb
+// The final gravitational potential energy is
+V_3p=W*h;// ft-lb
+v_B=sqrt(((T_1+V_1+U_13)-(V_3+V_3p))/0.828);// ft-lb
+printf("\n(a)The angular velocity of the bar,omega=%1.2f rad/sec \n(b)The velocity with which B strikes the horizontal surface,v_B=%1.2f ft/sec",omega,v_B);
diff --git a/3792/CH6/EX6.11/Ex6_11.sce b/3792/CH6/EX6.11/Ex6_11.sce new file mode 100644 index 000000000..42324eff5 --- /dev/null +++ b/3792/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,32 @@ +// SAMPLE PROBLEM 6/11
+clc;funcprot(0);
+// Given data
+m=30;// kg
+k=0.100;// m
+m_OB=10;// kg
+m_c=7;// kg
+K=30;// kN/m
+theta=45;// degree
+l=0.375;// m
+g=9.81;// m/s^2
+
+// Calculation
+// (a)
+// T_2=[2*((1/2)*I_G*omega^2]+[(1/2)*m*v^2];
+// T_2= 6.83*v_B^2;
+T_1=0;// J
+l_b=l/sqrt(2);// m
+V_1=(2*m_OB*g*(l_b/2))+(m_c*g*l_b);// J
+V_2=0;// J
+U_12=0;// J
+v_B=sqrt(((T_1+V_1+U_12)-(V_2))/6.83);// m/s
+// (b)
+T_3=0;// J
+U_13=0;// J
+function[X]=deformation(y)
+ X(1)=(T_1+V_1+U_13)-(T_3+((-2*m_OB*g*(y(1)/2))-(m_c*g*y(1))+((1/2)*K*10^3*y(1)^2)));
+endfunction
+y=[10];
+z=fsolve(y,deformation);
+x=z(1)*1000;// mm
+printf("\n(a)The velocity of the collar as it first strikes the spring,v_B=%1.2f m/s \n(b)The maximum deformation of the spring,x=%2.1f mm",v_B,x);
diff --git a/3792/CH6/EX6.12/Ex6_12.sce b/3792/CH6/EX6.12/Ex6_12.sce new file mode 100644 index 000000000..d76a77854 --- /dev/null +++ b/3792/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,19 @@ +// SAMPLE PROBLEM 6/12
+clc;funcprot(0);
+// Given data
+m_A=3;// kg
+m=2;// kg
+k=0.060;// The radius of gyration in m
+k=1.2;// The spring stiffness in kN/m
+F=80;// N
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// dT_rack=3a dx
+// dT_gear=0.781a dx
+// dV_rack=29.4 dx
+// dV_gear=9.81 dx
+// dV_spring=24 dx
+// Canceling dx and solving for a give
+a=(80-(29.4+9.81+24))/(3+0.781);
+printf("\nThe acceleration of rack A,a=%1.2f m/s^2",a);
diff --git a/3792/CH6/EX6.14/Ex6_14.sce b/3792/CH6/EX6.14/Ex6_14.sce new file mode 100644 index 000000000..fdcb83c0a --- /dev/null +++ b/3792/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,22 @@ +// SAMPLE PROBLEM 6/14
+clc;funcprot(0);
+// Given data
+// P=1.5*t;
+r_i=9/12;// ft
+r_o=18/12;// ft
+t_1=0;// s
+t_2=10;// s
+k=10/12;// ft
+W=120;// lb
+g=32.2;// The acceleration due to gravity in ft/sec^2
+v_1=-3;// ft/sec
+
+// Calculation
+function[X]=velocity(y)
+ X(1)=(((W/g)*v_1)+integrate('((1.5*t)-y(2))','t',t_1,t_2))-((W/g)*(r_o*y(1)));
+ X(2)=(((W/g)*(k)^2*(v_1/r_o))+integrate('((r_o*y(2))-(r_i*(1.5*t)))','t',t_1,t_2))-((W/g)*(k^2*y(1)));
+endfunction;
+y=[1 10];
+z=fsolve(y,velocity);
+omega_2=z(1);// rad/sec clockwise
+printf("\nThe angular velocity of the wheel,omega_2=%1.2f rad/sec",omega_2);
diff --git a/3792/CH6/EX6.15/Ex6_15.sce b/3792/CH6/EX6.15/Ex6_15.sce new file mode 100644 index 000000000..09024a579 --- /dev/null +++ b/3792/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,32 @@ +// SAMPLE PROBLEM 6/15
+clc;funcprot(0);
+// Given data
+m_E=30;// kg
+m_D=40;// kg
+v_1=1.2;// m/s
+t_1=0;// s
+t_2=5;// s
+F=380;// N
+d=375/1000;// m
+k_o=250/1000;// m
+g=9.81;// m/s^2
+
+// Calculation
+// [H_O1+(integral(t_2 to t_2))SigmaM_Odt=H_O2]
+// Integrating we get
+M=((((F*0.750)*t_2)-(((m_E+m_D)*g*d)*t_2))-(((F*0.750)*t_1)-(((m_E+m_D)*g*d)*t_1)));// N.m.s
+Ibar=(m_E)*k_o^2;// kg-m^2
+omega_1=v_1/d;//rad/sec
+H_O1=-((m_E+m_D)*v_1*d)-(Ibar*(v_1/d));// N.m.s
+// H_O2=-(m_E+m_D*v_2*d)-(Ibar*(v_2/d));
+// H_O2=11.72*omega_2;
+// Substituting into the momentum equation gives
+omega_2=(H_O1+M)/11.72;// N.m.s
+// [G_1+(integral(t_2 to t_2))SigmaFdt=G_2]
+m=m_E+m_D;// kg
+G_1=m*-(v_1);// (kg.m/s)
+G_2=m*(d*omega_2);// (kg.m/s)
+// Integrating
+// SigmaF=[T*(t_2)+(F*t_2)-(m*g*t_2)]-[T*(t_1)+(F*t_1)-(m*g*t_1)];
+T=((G_2-G_1)-(((F*t_2)-(m*g*t_2))-((F*t_1)-(m*g*t_1))))/(t_2-t_1);// N
+printf("\nThe angular velocity,omega_2=%1.2f rad/s counter clockwise \nThe tension in the cable,T=%3.0f N",omega_2,T);
diff --git a/3792/CH6/EX6.2/Ex6_2.sce b/3792/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..8c7597c4d --- /dev/null +++ b/3792/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,23 @@ +// SAMPLE PROBLEM 6/2
+clc;clear;funcprot(0);
+// Given data
+m=150;// kg
+M=5;// kN
+theta=30;// degree
+ACbar=1.5;// m
+BDbar=1.5;// m
+ABbar=1.8;// m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// SigmaM_C=0
+A_t=M/ACbar;// kN
+// SigmaF_t=m*abar_t
+// alpha=14.81-6.54*cos(theta);
+wsquare_30=(29.6*theta*%pi/180)-(13.08*sind(theta));// (rad/s)^2
+alpha_30=14.81-(6.54*cosd(theta));// rad/s^2
+A_n=(m/1000)*ACbar*wsquare_30;// kN
+A_t=(m/1000)*BDbar*alpha_30;// kN
+// SigmaM_A=m*abar*d
+B=((A_n*(ABbar-0.6)*cosd(theta))+(A_t*0.6))/(ABbar*cosd(theta));// kN
+printf("\nThe force in the link DB,B=%1.2f kN",B);
diff --git a/3792/CH6/EX6.3/Ex6_3.sce b/3792/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..720801322 --- /dev/null +++ b/3792/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,50 @@ +// SAMPLE PROBLEM 6/3
+clc;funcprot(0);
+// Given data
+W_b=644;// lb
+r_i=12;// inch
+r_o=24;// inch
+theta=45;// degree
+P=400;// lb
+k_o=18;// inch
+W=322;// lb
+g=32.2;// lb
+
+// Calculation
+// Solution 1
+// I=k^2*m
+Ibar=(k_o/12)^2*(W/g);// lb-ft-sec^2
+function[X]=acceleration(y)
+// SigmaM_G=Ibar*alpha
+ X(1)=((P*(r_o/12))-(y(1)*(r_i/12)))-(Ibar*y(2));
+// SigmaF_y=m*a_y
+ X(2)=((y(1)-W_b))-((W_b/g)*y(3));
+// a_t=r*a;
+ X(3)=y(3)-((r_i/12)*y(2));
+endfunction
+y=[100 1 1];
+z=fsolve(y,acceleration);
+T=z(1);// lb
+alpha=z(2);// rad/sec^2
+a=z(3);// ft/sec^2
+// SigmaF_x=0
+O_x=P*cosd(theta);// lb
+// SigmaF_y=0
+O_y=W+T+(P*sind(theta));// lb
+O=sqrt(O_x^2+O_y^2);// lb
+printf("\nSolution I:T=%3.0f lb,alpha=%1.2f rad/sec^2,a=%1.2f ft/sec^2,O=%4.0f lb",T,alpha,a,O);
+// Solution 2
+function[Y]=acceleration(x)
+// SigmaM_o=(Ibar*alpha)+(m*abar*d)
+ Y(1)=((P*(r_o/12))-(W_b*(r_i/12)))-((Ibar*x(1))+((W_b/g)*x(2)*(r_i/12)));
+// a_t=r*a;
+ Y(2)=x(2)-((r_i/12)*x(1));
+endfunction
+x=[1 1];
+m=fsolve(x,acceleration);
+alpha=m(1);// rad/sec^2
+a=m(2);// ft/sec^2
+// SigmaF_y=Sigmam*(a_ybar)
+O_y=(W+W_b+(P*sind(theta)))+(((W/g)*(0))+((W_b/g)*alpha));// lb
+// SigmaF_x=Sigmam*(a_xbar)
+O_x=P*sind(theta);// lb
diff --git a/3792/CH6/EX6.4/Ex6_4.sce b/3792/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..2446945df --- /dev/null +++ b/3792/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,23 @@ +// SAMPLE PROBLEM 6/4
+clc;funcprot(0);
+// Given data
+m=7.5;// kg
+rbar=250/1000;// m
+k_o=295/1000;// m
+theta_1=0;// degree
+theta_2=60;// degree
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// SigmaM_o=I_o*alpha;
+// alpha=28.2*cos(theta);
+wsquare=48.8;// (rad/s)^2
+// SigmaF_n=m*rbar*omega^2;
+O_n=(m*rbar*wsquare)+(m*g*sind(theta_2));// N
+// SigmaF_t=m*rbar*alpha;
+O_t=(m*g*cosd(theta_2))-(m*rbar*28.2*cosd(theta_2));// N
+O=sqrt(O_n^2+O_t^2);// N
+q=k_o^2/(rbar);// The distance in m
+// SigmaM_Q=0
+O_t=(m*g*cosd(theta_2)*(q-rbar))/q;// N
+printf("\nThe total force supported by the bearing,O=%3.1f N \nO_t=%2.2f N",O,O_t);
diff --git a/3792/CH6/EX6.5/Ex6_5.sce b/3792/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..104694086 --- /dev/null +++ b/3792/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,26 @@ +// SAMPLE PROBLEM 6/5
+clc;funcprot(0);
+// Given data
+r=6/12;// ft
+mu_s=0.15;// The coefficients of static friction
+mu_k=0.12;// The coefficients of kinetic friction
+theta=20;// degree
+g=32.2;// The acceleration due to gravity in ft/sec^2
+x=10;// ft
+
+// Calculation
+// SigmaF_x=m*abar_x----> mg*sind(theta)-F=m*abar
+// SigmaF_x=m*abar_y----> N-mg*cosd(theta)=0
+// SigmaM_G=Ibar*alpha---> F*r=m*r^2*alpha
+abar=(g/2)*sind(theta);// ft/sec^2
+// SigmaM_G=Ibar*alpha+m*abar*d----->mgr*sin(theta)=mr^2*(abar/r)+ m*abar*r
+// From the above equations,we solve using the coefficients of mg
+F=sind(theta)-(sind(theta))/2;// N
+N=cosd(theta);// N
+F_max=mu_s*N;// N
+F=mu_k*N;// N
+// SigmaF_x=m*abar_x
+abar=(sind(theta)-F)*g;// ft/sec^2
+alpha=(F*g)/r;// rad/sec^2
+t=sqrt((2*x)/abar);// sec
+printf("\nThe angular acceleration of the hoop,alpha=%1.2f ft/sec^2 \nThe time t for the hoop to move a distance of 10 ft down the incline,t=%1.3f sec",alpha,t);
diff --git a/3792/CH6/EX6.6/Ex6_6.sce b/3792/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..05d165655 --- /dev/null +++ b/3792/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,37 @@ +// SAMPLE PROBLEM 6/6
+clc;funcprot(0);
+// Given data
+alpha_0=3;// rad/s^2
+m=70;// kg
+k=0.250;// The radius of gyration in m
+mu_s=0.25;// The coefficient of static friction
+g=9.81;// The acceleration due to gravity in m/s^2
+DCbar=0.30;// m
+r_A=0.250;// m
+r_Bi=0.150;// m
+r_Bo=0.450;// m
+
+// Calculation
+a_t=r_A*alpha_0;// m/s^2
+alpha=a_t/DCbar;// rad/s^2
+abar=r_Bo*alpha;// m/s^2
+function[X]=force(y)
+ // SigmaF_x=m*abar_x
+ X(1)=(y(1)-y(2))-(m*-abar);
+ N=(m*g);// N
+ // SigmaM_G=Ibar*alpha
+ X(2)=((r_Bo*y(1))-(r_Bi*y(2)))-(m*k^2*alpha);
+endfunction
+y=[10 100];
+z=fsolve(y,force);
+F=z(1);// N
+T=z(2);// N
+printf("\nThe tension in the cable,T=%3.1f N \nThe friction force exerted by the horizontal surface on the spool,F=%2.1f N",T,F);
+N=(m*g);// N
+F_max=mu_s*N;// N
+// If the coefficient of static friction had been 0.1
+mu_s=0.1;// The coefficient of static friction
+F=mu_s*(m*g);// N
+// SigmaM_C=Ibar*alpha + m*abar*r
+T=((m*(r_A^2)*alpha)+(m*abar*r_Bo))/DCbar;// N
+printf("\nThe tension in the cable,T=%3.1f N",T);
diff --git a/3792/CH6/EX6.7/Ex6_7.sce b/3792/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..01def36a9 --- /dev/null +++ b/3792/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,38 @@ +// SAMPLE PROBLEM 6/7
+clc;funcprot(0);
+// Given data
+W=60;// lb
+theta=30;// degree
+F=30;// lb
+BGbar=2;// ft
+AGbar=2;// ft
+l=4;// ft
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+// abar_x=abar*cos(theta)=1.732*alpha;
+// abar_y=abar*sin(theta)=1.0*alpha;
+function[X]=force(y)
+ // SigmaM_G=Ibar*alpha;
+ X(1)=((F*(2*cosd(theta)))-(y(1)*(AGbar*sind(theta)))+(y(2)*(BGbar*cosd(theta))))-((1/12)*(W/g)*l^2*y(3));
+ // SigmaF_x=m*abar_x;
+ X(2)=(F-y(2))-((W/g)*(2*cosd(theta)*y(3)));
+ // SigmaF_y=m*abar_y;
+ X(3)=(y(1)-W)-((W/g)*2*sind(theta)*y(3));
+endfunction
+y=[10 10 1];
+z=fsolve(y,force);
+A=z(1);// lb
+B=z(2);// lb
+alpha=z(3);// rad/sec^2
+printf("\nThe forces on the small end rollers ,A=%2.1f lb and B=%2.2f lb \nThe resulting angular acceleration of the bar,alpha=%1.2f rad/sec^2",A,B,alpha);
+// Alternative solution
+// SigmaM_C=(Ibar*alpha)+(Sigma m*abar*d)
+alpha=((F*(l*cosd(theta)))-(W*2*sind(theta)))/(((1/12)*(W/g)*l^2)+((W/g)*1.732*2*cosd(theta))+((W/g)*1*2*sind(theta)));// rad/sec^2
+// SigmaF_x=m*abar_x;
+abar_y=2*alpha*sind(theta);// ft
+A=((W/g)*abar_y)+W;// lb
+// SigmaF_x=m*abar_x;
+abar_x=2*alpha*cosd(theta);// ft
+B=F-((W/g)*abar_x);// lb
+printf("\nAlternative solution: \nThe forces on the small end rollers ,A=%2.1f lb and B=%2.2f lb \nThe resulting angular acceleration of the bar,alpha=%1.2f rad/sec^2",A,B,alpha);
diff --git a/3792/CH6/EX6.9/Ex6_9.sce b/3792/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..753318c96 --- /dev/null +++ b/3792/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,24 @@ +// SAMPLE PROBLEM 6/9
+clc;funcprot(0);
+// Given data
+F=100;// N
+m=40;// kg
+k=0.150;// m
+theta=15;// degree
+r_i=0.100;// m
+r_o=0.200;// m
+l=3;// The distance in m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+W=m*g;// N
+l=(r_o+r_i)/r_i;// m
+U_12=(F*((r_o+r_i)/r_i)*l)-((W*sind(theta)*l));// J
+T_1=0;// J
+// T_2=((1/2)*m*vbar^2)+((1/2)*Ibar*omega^2);
+// The work-energy equation gives
+omega=sqrt((T_1+U_12)/(((1/2)*m*(r_i)^2)+((1/2)*m*k^2)));// rad/s
+// Alternatively, the kinetic energy of the wheel may be written
+// T=(1/2)*I_C*omega^2
+P_100=F*(r_o+r_i)*omega;// W
+printf("The power input,P=%3.0f W",P_100);
diff --git a/3792/CH7/EX7.1/Ex7_1.sce b/3792/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..e4ca762ce --- /dev/null +++ b/3792/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,34 @@ +// SAMPLE PROBLEM 7/1
+clc;funcprot(0);
+// Given data
+L=0.8;// m
+N=60;// rev/min
+betadot=4;// rad/s
+beta=30;// degree
+
+// Solution
+// (a)
+omega_x=betadot;// (i) rad/s
+omega_z=(2*%pi*N/60);// (k) rad/s
+omega=[omega_x,0,omega_z];// (i,j,k) rad/s
+printf("\n(a)The angular velocity of OA,omega=%1.0fi+%1.2fk rad/s",omega(1),omega(3));
+// (b)
+omegadot_z=0;// (k) rad/s
+omegadot_x=omega_z*omega_x;// (i) rad/s
+alpha=omegadot_x+omegadot_z;// (j) rad/s^2
+alpha=[0,alpha,0];// (i,j,k) rad/s^2
+printf("\n(b)The angular acceleration of OA,alpha=%2.1fj rad/s^2",alpha(2));
+// (c)
+r=[0,0.693,0.4];// m
+// v=omega*r;
+v_1=det([omega(2),omega(3);r(2),r(3)]);// m/s
+v_2=-det([omega(1),omega(3);r(1),r(3)]);// m/s
+v_3=det([omega(1),omega(2);r(1),r(2)]);// m/s
+v=[v_1,v_2,v_3];// m/s
+printf("\n(c)The velocity of point A,v=%1.2fi+(%1.2f)j+%1.2fk m/s",v(1),v(2),v(3));
+// (d)
+a_1=det([alpha(2),alpha(3);r(2),r(3)])+det([omega(2),omega(3);v(2),v(3)]);// m/s^2
+a_2=-det([alpha(1),alpha(3);r(1),r(3)])+(-det([omega(1),omega(3);v(1),v(3)]));// m/s^2
+a_3=det([alpha(1),alpha(2);r(1),r(2)])+det([omega(1),omega(2);v(1),v(2)]);// m/s^2
+a=[a_1,a_2,a_3];// m/s^2
+printf("\n(d)The acceleration of point A,v=%2.1fi+(%2.1f)j+(%1.2f)k m/s^2",a(1),a(2),a(3));
diff --git a/3792/CH7/EX7.2/Ex7_2.sce b/3792/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..f2685b098 --- /dev/null +++ b/3792/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,32 @@ +// SAMPLE PROBLEM 7/2
+clc;funcprot(0);
+// Given data
+N_0=120;// rev/min
+N=60;// rev/min
+gamma=30;// degree
+OCbar=10;// inch
+CAbar=5;// inch
+theta=30;// degree
+
+// Calculation
+// (a)
+omega_0=(2*%pi*N_0)/60;// rad/sec
+omega_1=(2*%pi*N)/60;// rad/sec
+omega=[0,(omega_1*cosd(gamma)),(omega_0+(omega_1*sind(theta)))];// rad/sec
+printf("\n(a)The angular velocity,omega=%1.2fj+%2.2fk rad/s",omega(2),omega(3));
+alpha=[(omega_1*omega_0*cosd(theta)),0,0];// rad/sec^2
+printf("\n(b)The angular acceleration,alpha=%2.1fi rad/s^2",alpha(1));
+r=[0,5,10];// inch
+// (c)
+// v=omega*r;
+v_1=det([omega(2),omega(3);r(2),r(3)]);// in/sec
+v_2=-det([omega(1),omega(3);r(1),r(3)]);// in/sec
+v_3=det([omega(1),omega(2);r(1),r(2)]);// in/sec
+v=[v_1,v_2,v_3];// in/sec
+printf("\n(c)The velocity of point A,v=%2.1fi+(%1.0f)j+%1.fk in/sec",v(1),v(2),v(3));
+// a=(alpha*r)+(omega*v)
+a_1=det([alpha(2),alpha(3);r(2),r(3)])+det([omega(2),omega(3);v(2),v(3)]);// in/sec^2
+a_2=-det([alpha(1),alpha(3);r(1),r(3)])+(-det([omega(1),omega(3);v(1),v(3)]));// in/sec^2
+a_3=det([alpha(1),alpha(2);r(1),r(2)])+det([omega(1),omega(2);v(1),v(2)]);// in/sec^2
+a=[a_1,a_2,a_3];// in/sec^2
+printf("\n The acceleration of point A,a=%1.0fi+(%1.0f)j+%3.0fk in/sec^2",a(1),a(2),a(3));
diff --git a/3792/CH7/EX7.3/Ex7_3.sce b/3792/CH7/EX7.3/Ex7_3.sce new file mode 100644 index 000000000..d05d17cb4 --- /dev/null +++ b/3792/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,29 @@ +// SAMPLE PROBLEM 7/3
+clc;funcprot(0);
+// Given data
+omega_1=6;// rad/s
+r_x=50;// mm
+r_y=100;// mm
+r_z=100;// mm
+
+
+// Calculation
+// v_A=r_x*omega_2;
+v_B=r_y*omega_1;// (i) mm/s
+// v_A=v_B+(omega_n*r_A/B);
+// Expanding the determinant and equating the coefficients of the i, j, k terms give
+function[X]=velocity(y)
+ X(1)=-6-(y(2)-y(3));
+ X(2)=y(4)-((-2*y(1))+y(3));
+ X(3)=0-((2*y(1))-y(2));
+ X(4)=((r_x*y(1))+(r_y*y(2))+(r_z*y(3)));
+endfunction
+y=[1 1 1 1];
+z=fsolve(y,velocity);
+omega_nx=z(1);// rad/s
+omega_ny=z(2);// rad/s
+omega_nz=z(3);// rad/s
+omega_2=z(4);// rad/s
+omega_n=[omega_nx,omega_ny,omega_nz];// rad/s
+omega_n=norm(omega_n);// rad/s
+printf("\nThe angular velocity of crank DA,omega_2=%1.0f rad/s \nThe angular velocity of link AB,omega_n=%1.3f rad/s",omega_2,omega_n);
diff --git a/3792/CH7/EX7.4/Ex7_4.sce b/3792/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..73ad45fd9 --- /dev/null +++ b/3792/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,33 @@ +// SAMPLE PROBLEM 7/4
+clc;funcprot(0);
+// Given data
+// From sample problem 7/3
+omega_1=6;// rad/s
+omega_2=6;// rad/s
+r_x=50;// mm
+r_y=100;// mm
+r_z=100;// mm
+omega_n=2*sqrt(5);// rad/s
+
+// Calculation
+r_AB=[r_x,r_y,r_z];// mm
+// a_A=[r_x*omega_2^2]i+[r_x*omegadot]j;
+// a_B=[r_y*omega_1^2]k+[0]i;
+omegadot=(omega_n)^2*(r_AB);// rad/s^2
+// omegadot*r_A/B=(100*omegadot_ny-100*omegadot_nz)i+(50*omegadot_nz-100*omegadot_nx)j+(100*omegadot_nx-50omegadot_ny)k
+function[X]=velocity(y)
+ X(1)=28-(y(2)-y(3));
+ X(2)=(y(4)+40)-((-2*y(1))+y(3));
+ X(3)=-32-((2*y(1))-y(2));
+ X(4)=((2*y(1))+(4*y(2))+(4*y(3)));
+endfunction
+y=[1 10 10 10];
+z=fsolve(y,velocity);
+omegadot_nx=z(1);// rad/s^2
+omegadot_ny=z(2);// rad/s^2
+omegadot_nz=z(3);// rad/s^2
+omegadot_2=z(4);// rad/s^2
+omegadot_n=[omegadot_nx,omegadot_ny,omegadot_nz];// rad/s^2
+omegadot_n=norm(omegadot_n);// rad/s^2
+printf("\nThe angular acceleration of crank AD,omegadot_2=%2.0f rad/s \nThe angular acceleration of link AB,omegadot_n=%2.2f rad/s",omegadot_2,omegadot_n);
+
diff --git a/3792/CH7/EX7.5/Ex7_5.sce b/3792/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..a098ea34c --- /dev/null +++ b/3792/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,51 @@ +// SAMPLE Pr_BOBLEM 7/5
+clc;funcprot(0);
+// Given data
+omega=3;// rad/s
+p=8;// rad/s
+gamma=30;// degree
+y=0.300;// m
+z=0.120;// m
+
+// Calculation
+// Velocity
+omega=[0,0,3];// rad/s
+r_B=[0,0.350,0];// m
+v_B1=det([omega(2),omega(3);r_B(2),r_B(3)]);// m/s
+v_B2=-det([omega(1),omega(3);r_B(1),r_B(3)]);// m/s
+v_B3=det([omega(1),omega(2);r_B(1),r_B(2)]);// m/s
+v_B=[v_B1,v_B2,v_B3];// m/s
+// Note that k*i=J=jcos(gamma)-ksin(gamma),K*j=-i*cos(gamma) and K*k=i*sin(gamma)
+r_AB=[0,y,z];// m
+// omega*r_AB=3K*(yj+zk);
+omegaintor_AB=(-(omega(3)*(y*cosd(gamma))))+(omega(3)*(z*sind(gamma)));// m/s
+p=[0,8,0];// rad/s
+v_rel1=det([p(2),p(3);r_AB(2),r_AB(3)]);// m/s
+v_rel2=-det([p(1),p(3);r_AB(1),r_AB(3)]);// m/s
+v_rel3=det([p(1),p(2);r_AB(1),r_AB(2)]);// m/s
+v_rel=[v_rel1,v_rel2,v_rel3];// m/s
+v_A=v_B(1)+omegaintor_AB+v_rel(1);// m/s
+printf("\nThe velocity of point A,v_A=%0.4fi m/s",v_A);
+// Acceleration
+a_B1=det([omega(2),omega(3);v_B(2),v_B(3)]);// m/s^2
+a_B2=-det([omega(1),omega(3);v_B(1),v_B(3)]);// m/s^2
+a_B3=det([omega(1),omega(2);v_B(1),v_B(2)]);// m/s^2
+a_B=[a_B1,a_B2,a_B3];// m/s^2
+a_B=[0,((a_B(2)*(cosd(gamma)))),-(a_B(2)*(sind(gamma)))];// m/s^2
+omegadot=0;// m/s^2
+// Assume O=omega*(omega*r_A/B)
+O=[0,((omega(3)*omegaintor_AB*(cosd(gamma)))),-omega(3)*(omegaintor_AB*(sind(gamma)))];// m/s^2
+// Assume O_1=2*omega*v_rel
+O_1=[0,((2*omega(3)*v_rel(1)*(cosd(gamma)))),-2*omega(3)*(v_rel(1)*(sind(gamma)))];// m/s^2
+a_rel1=det([p(2),p(3);v_rel(2),v_rel(3)]);// m/s^2
+a_rel2=-det([p(1),p(3);v_rel(1),v_rel(3)]);// m/s^2
+a_rel3=det([p(1),p(2);v_rel(1),v_rel(2)]);// m/s^2
+a_rel=[a_rel1,a_rel2,a_rel3];// m/s^2
+a_A=[(a_B(1)+(omegadot*r_AB(1))+O(1)+O_1(1)+a_rel1),(a_B(2)+(omegadot*r_AB(2))+O(2)+O_1(2)+a_rel2),(a_B(3)+(omegadot*r_AB(3))+O(3)+O_1(3)+a_rel3)];// m/s^2
+a_A=norm(a_A);// m/s^2
+printf("\nThe acceleration of point A,a_A=%1.2f m/s",a_A);
+// Angular Acceleration
+// Note that k*i=J=jcos(gamma)-ksin(gamma),K*j=-i*cos(gamma) and K*k=i*sin(gamma)
+omega=[3,8];// rad/s (K,j)(k*j=-i*cos(gamma))
+alpha=[0+(-omega(1)*omega(2)*cosd(gamma))];// (i) rad/s^2
+printf("\nThe angular acceleration of the disk,alpha=%2.1fi rad/s^2",alpha);
diff --git a/3792/CH7/EX7.6/Ex7_6.sce b/3792/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..5a6ce6d19 --- /dev/null +++ b/3792/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,41 @@ +// SAMPLE PROBLEM 7/6
+clc;funcprot(0);
+// Given data
+m=70;// The mass of bent plate in kg
+omega=30;// rad/s
+x_A=0.125;// m
+y_A=0.100;// m
+x_B=0.075;// m
+y_B=.150;// m
+d_x=0.0375;// m
+d_y=0.125;// m
+d_z=0.075;// m
+
+// Calculation
+// Part A
+m_A=x_A*y_A*m;// kg
+m_B=x_B*y_B*m;// kg
+I_xxA=((m_A/12)*(y_A^2+x_A^2))+(m_A*((x_A/2)^2+(y_A/2)^2));// kg.m^2
+I_yyA=(m_A/3)*(y_A)^2;// kg.m^2
+I_zzA=(m_A/3)*(x_A)^2;// kg.m^2
+I_xyA=0;// kg.m^2
+I_xzA=0;// kg.m^2
+I_yzA=0+(m_A*(x_A/2)*(y_A/2));// kg.m^2
+// Part B
+I_xxB=((m_B/12)*(y_B^2))+((m_B)*(d_y^2+d_z^2));// kg.m^2
+I_yyB=((m_B/12)*(x_B^2+y_B^2))+(m_B*(d_x^2+d_z^2));// kg.m^2
+I_zzB=((m_B/12)*(x_B^2))+(m_B*((x_A)^2+(d_x^2)));// kg.m^2
+I_xyB=0+(m_B*d_x*d_y);// kg.m^2
+I_xzB=0+(m_B*d_x*d_z);// kg.m^2
+I_yzB=0+(m_B*d_y*d_z);// kg.m^2
+I_xx=I_xxA+I_xxB;// kg.m^2
+I_yy=I_yyA+I_yyB;// kg.m^2
+I_zz=I_zzA+I_zzB;// kg.m^2
+I_xy=I_xyA+I_xyB;// kg.m^2
+I_xz=I_xzA+I_xzB;// kg.m^2
+I_yz=I_yzA+I_yzB;// kg.m^2
+// (a)
+H_o=[-(omega*I_xz),-(omega*I_yz),(omega*I_zz)];// The angular momentum of the body in N.m.s
+// (b)
+T=(1/2)*(omega)*[H_o(3)];//(k.i=0,k.j=0,k.k=1) The kinetic energy in J
+printf("\n(a)The angular momentum H of the plate about point O,H_O=%0.4fi+(%0.4f)j+%0.4fk \n(b)The kinetic energy of the plate,T=%1.2f J",H_o(1),H_o(2),H_o(3),T);
diff --git a/3792/CH7/EX7.8/Ex7_8.sce b/3792/CH7/EX7.8/Ex7_8.sce new file mode 100644 index 000000000..230175198 --- /dev/null +++ b/3792/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,25 @@ +// SAMPLE PROBLEM 7/8
+clc;funcprot(0);
+// Given data
+m=1000;// The mass of turbine rotor in kg
+k=0.200;// m
+N=500;// rev/min
+rho=400;// The radius of gyration in m
+v=25*0.514;// m/s
+d_AG=0.6;// m
+d_GB=0.9;// m
+d_AB=d_AG+d_GB;// m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// The moment principle from statics easily gives
+W=m*g;// N
+R_1=(m*g)*d_AG;// N
+R_2=W-R_1;// N
+omega=(v/rho);// rad/s
+I=m*k^2;// kg-m^2
+deltaR=(I*omega*((2*%pi*N)/60))/d_AB;
+R_A=R_1-deltaR;// N
+R_B=R_2+deltaR;// N
+printf("\nThe vertical components of the bearing reactions at A and B,R_A=%4.0f N and R_B=%4.0f N",R_A,R_B);
+// The answer provided in the textbook is wrong
diff --git a/3792/CH7/EX7.9/Ex7_9.sce b/3792/CH7/EX7.9/Ex7_9.sce new file mode 100644 index 000000000..fae018e14 --- /dev/null +++ b/3792/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,20 @@ +// SAMPLE PROBLEM 7/9
+clc;funcprot(0);
+// Given data
+t=4;// s
+theta=20;// degree
+p=(2*%pi)/4;// rad/s
+
+// Calculation
+// (a)
+// I_zz=(56/3)*mr^2;
+// I_xx=(32/3)*mr^2;
+// By using coefficient of I_xx and I_zz
+I=56/3;// The moment of inertia
+I_0=32/3;// The moment of inertia
+costheta=1;// radian
+n=I/((I_0-I)*costheta);// The ratio of angular rates
+// (b)
+tau=((2*%pi)/p)*abs(((I_0-I)/I)*cosd(theta));// s
+beta=atand((I/I_0)*tand(theta));// degree
+printf("\n(a)The number of complete cycles,n=%1.2f \n The minus sign indicates retrograde precession where, in the present case,and p are essentially of opposite sense. Thus, the station will make seven wobbles for every three revolutions. \n(b)The period of precession,tau=%1.3f s",n,tau);
diff --git a/3792/CH8/EX8.1/Ex8_1.sce b/3792/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..ca567445f --- /dev/null +++ b/3792/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,18 @@ +// SAMPLE PROBLEM 8/1
+clc;funcprot(0);
+// Given data
+W=25;// The weight of the body in lb
+k=160;// lb/ft
+v=2;// The downward velocity in ft/sec
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+// Calculation
+// (a)
+delta_st=W/k;// The static spring deflection in ft
+delta_st=delta_st*12;// in
+// (b)
+omega_n=sqrt(k/(W/g));// The natural frequency of the system in rad/sec
+f_n=omega_n*(1/(2*%pi));// The natural frequency of the system in cycles/sec
+// (c)
+tau=1/f_n;// The system period in sec
+printf("\n(a)The static spring deflection,delta_st=%0.4f ft (or)%1.3f in \n(b)The natural frequency of the system,omega_n=%2.2f rad/sec \n The natural frequency of the system,f_n=%0.3f sec \n(c)The system period,tau=%0.3f sec",delta_st/12,delta_st,omega_n,f_n,tau);
diff --git a/3792/CH8/EX8.11/Ex8_11.sce b/3792/CH8/EX8.11/Ex8_11.sce new file mode 100644 index 000000000..3df59a453 --- /dev/null +++ b/3792/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,13 @@ +// SAMPLE PROBLEM 8/11
+clc;funcprot(0);
+// Given data
+m_c=3;// The mass of collar in kg
+m_l=1.2;// The mass of the links in kg
+k=1.5;// The stiffness of the spring in kN/m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+P=(m_c*g)+(2*(1/2)*m_l*g);// The compression P in N
+delta_st=P/(k*10^3);// The static deflection of the spring in m
+omega_n=sqrt(750/1.9);// Hz;
+printf("\nThe natural frequency of vertical vibration,omega_n=%2.2f Hz",omega_n);
diff --git a/3792/CH8/EX8.2/Ex8_2.sce b/3792/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..0776e6945 --- /dev/null +++ b/3792/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,16 @@ +// SAMPLE PROBLEM 8/2
+clc;funcprot(0);
+// Given data
+m=8;// kg
+s=0.2;// m
+t_1=0;// s
+t_2=2;// s
+c=20;// N.s/m
+k=32;// N/m
+
+// Calculation
+omega_n=sqrt(k/m);// rad/s
+eta=c/(2*m*omega_n);// The damping ratio
+omega_d=omega_n*(sqrt(1-eta^2));// The damped natural frequency in rad/s
+x_2=0.256*(exp((-1.25*t_2)))*(sin((1.561*t_2)+0.896));// m
+printf("\nThe displacement in meters, x_2=%0.5f m",x_2);
diff --git a/3792/CH8/EX8.4/Ex8_4.sce b/3792/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..ddf406946 --- /dev/null +++ b/3792/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,14 @@ +// SAMPLE PROBLEM 8/4
+clc;funcprot(0);
+// Given data
+m=50;// kg
+n=4;// Number of springs
+// x_B=0.002cos50t
+b=0.002;// m
+omega=50;// rad/s
+k=7500;// The stiffness of the spring in N/m
+
+// Calculation
+omega_n=sqrt((n*k)/m);// The resonant frequency in rad/s
+X=b/(1-(omega/omega_n)^2);// m
+printf("\nThe amplitude of the steady-state motion of the instrument,X=%1.2e m (or) %0.3f mm",X,X*10^3);
diff --git a/3792/CH8/EX8.6/Ex8_6.sce b/3792/CH8/EX8.6/Ex8_6.sce new file mode 100644 index 000000000..f4db86a9e --- /dev/null +++ b/3792/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,20 @@ +// SAMPLE PROBLEM 8/6
+clc;funcprot(0);
+// Given data
+W=100;// The weight of the piston in lb
+k=200;// The spring modulus in lb/in
+c=85;// The damping coefficient in lb-sec/ft
+a=80;// The top surface area in in^2
+omega=30;// rad/s
+g=32.2;// The acceleration due to gravity in ft/sec^2
+p=0.625;// lb/in^2
+
+// Calculation
+omega_n=sqrt((k*12)/(W/g));// The natural frequency of the system in rad/sec
+eta=c/(2*(W/g)*omega_n);// The damping ratio
+F_0=p*a;// lb
+X=(F_0/(k*12))/((1-(omega/omega_n)^2)^2+(2*eta*omega/omega_n)^2)^(1/2);// The steady-state amplitude in ft
+phi=atan((2*eta*omega/omega_n)/(1-(omega/omega_n)^2));// The phase angle in rad
+// x_p=Xsin(omega*t-phi);
+F_trmax=X*sqrt((k*12)^2+(c^2*omega^2));// The maximum force transmitted to the base in lb
+printf("\nThe steady-state displacement as a function of time,x_p=%0.5fsin(%2.0ft-(%1.3f))ft \nThe maximum force transmitted to the base,(F_tr)_max=%2.1f lb",X,omega,phi,F_trmax);
diff --git a/3792/CH8/EX8.7/Ex8_7.sce b/3792/CH8/EX8.7/Ex8_7.sce new file mode 100644 index 000000000..4c0636b10 --- /dev/null +++ b/3792/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,10 @@ +// SAMPLE PROBLEM 8/7
+clc;funcprot(0);
+// Given data
+rbar=0.9;// m
+k_o=0.95;// The radius of gyration in m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+tau=2*%pi*sqrt(k_o^2/(g*rbar));// The period for small oscillations about the pivot in s
+printf("\nThe period for small oscillations about the pivot,tau=%1.2f s",tau);
diff --git a/3792/CH8/EX8.9/Ex8_9.sce b/3792/CH8/EX8.9/Ex8_9.sce new file mode 100644 index 000000000..7dd8ff582 --- /dev/null +++ b/3792/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,25 @@ +// SAMPLE PROBLEM 8/9
+clc;funcprot(0);
+// Given data
+m=50;// The mass of the cylinder in kg
+r=0.5;// The cylinder radius in m
+k=75;// The spring constant in N/m
+c=10;// The damping coefficient in N.s/m
+x=-0.2;// m
+t=0;// s
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+omega_n=sqrt((2/3)*(k/m));// The undamped natural frequency in rad/s
+eta=(1/3)*(c/(m*omega_n));// The damping ratio
+omega_d=omega_n*(sqrt(1-eta^2));// The damped natural frequency in rad/s
+tau_d=(2*%pi)/omega_d;// The period of the damped system in s
+function[X]=Candpsi(y)
+ X(1)=(y(1)*sin(y(2)))-(-0.2);
+ X(2)=((-0.0667*y(1)*sin(y(2)))+((0.998*y(1)*cos(y(2)))))-0;
+endfunction
+y=[0.1 1.1];
+z=fsolve(y,Candpsi);
+C=z(1);// m
+psi=z(2);// rad
+printf("\n(a)The undamped natural frequency,omega_n=%1.0f rad/s \n(b)The damping ratio,eta=%0.4f \n(c)The damped natural frequency,omega_d=%0.3f rad/s \n(d)The period of the damped system,tau=%1.2f s \nThus, the motion is given by x=%0.3fexp(-%0.4f*t)sin(%0.3ft+%1.3f)m",omega_n,eta,omega_d,tau_d,C,eta,omega_d,psi);
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