diff options
Diffstat (limited to '3792/CH3')
26 files changed, 480 insertions, 0 deletions
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);
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