diff options
Diffstat (limited to '3886/CH2')
44 files changed, 448 insertions, 0 deletions
diff --git a/3886/CH2/EX2.1/2_1.txt b/3886/CH2/EX2.1/2_1.txt new file mode 100644 index 000000000..a27271764 --- /dev/null +++ b/3886/CH2/EX2.1/2_1.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_1.sce', -1)
+Magnitude of forces are :-
+ F1=100 N
+ F2=200 N
+Angle between the forces is :-
+ theta=63.9 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.10/2_10.txt b/3886/CH2/EX2.10/2_10.txt new file mode 100644 index 000000000..1187f80da --- /dev/null +++ b/3886/CH2/EX2.10/2_10.txt @@ -0,0 +1,19 @@ +
+
+Startup execution:
+ loading initial environment
+--> //Determining horizontal force F
+--> //From fig 2.14(b)
+--> //Resolving the forces
+--> //Fy=0 gives
+--> R=1500/cosd(30) //N
+ R =
+
+ 1732.0508
+--> //Fx=0 gives
+--> F=R*sind(30) //N
+ F =
+
+ 866.0254
+--> printf("Horizontal force of F=%.0f N is required to be applied",F)
+Horizontal force of F=866 N is required to be applied
\ No newline at end of file diff --git a/3886/CH2/EX2.10/Ex2_10.sce b/3886/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..13b004ff6 --- /dev/null +++ b/3886/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,9 @@ +//Determining horizontal force F
+//From fig 2.14(b)
+//Resolving the forces
+//Fy=0 gives
+R=1500/cosd(30) //N
+//Fx=0 gives
+F=R*sind(30) //N
+printf("Horizontal force of F=%.0f N is required to be applied",F)
+
diff --git a/3886/CH2/EX2.11/2_11.txt b/3886/CH2/EX2.11/2_11.txt new file mode 100644 index 000000000..2b8da0e66 --- /dev/null +++ b/3886/CH2/EX2.11/2_11.txt @@ -0,0 +1,14 @@ + --> //Finding forces developed in wires
+--> //applying Lami's theorem
+--> T1=150*sind(90+60)/sind(45+30) //N
+ T1 =
+
+ 77.645714
+--> T2=150*sind(180-45)/sind(45+30) //N
+ T2 =
+
+ 109.80762
+--> printf("The forces in the wires are:-\nT1=%.1f N \nT2=%.1f N",T1,T2)
+The forces in the wires are:-
+T1=77.6 N
+T2=109.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.11/Ex2_11.sce b/3886/CH2/EX2.11/Ex2_11.sce new file mode 100644 index 000000000..033ee5920 --- /dev/null +++ b/3886/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,6 @@ +//Finding forces developed in wires
+//applying Lami's theorem
+T1=150*sind(90+60)/sind(45+30) //N
+T2=150*sind(180-45)/sind(45+30) //N
+printf("The forces in the wires are:-\nT1=%.1f N \nT2=%.1f N",T1,T2)
+
diff --git a/3886/CH2/EX2.12/2_12.txt b/3886/CH2/EX2.12/2_12.txt new file mode 100644 index 000000000..5dc08eb40 --- /dev/null +++ b/3886/CH2/EX2.12/2_12.txt @@ -0,0 +1,15 @@ + --> //Determine reactions at contact
+--> //Refer fig.2.16(b)
+--> //applying Lami's Theorem
+--> R1=400*sind(180-45)/sind(60+45) //N
+ R1 =
+
+ 292.82032
+--> R2=400*sind(180-60)/sind(60+45) //N
+ R2 =
+
+ 358.63019
+--> printf("The reactions developed are:-\nR1=%.1f N \nR2=%.1f N",R1,R2)
+The reactions developed are:-
+R1=292.8 N
+R2=358.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.12/Ex2_12.sce b/3886/CH2/EX2.12/Ex2_12.sce new file mode 100644 index 000000000..c3f8a027f --- /dev/null +++ b/3886/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,6 @@ +//Determine reactions at contact
+//Refer fig.2.16(b)
+//applying Lami's Theorem
+R1=400*sind(180-45)/sind(60+45) //N
+R2=400*sind(180-60)/sind(60+45) //N
+printf("The reactions developed are:-\nR1=%.1f N \nR2=%.1f N",R1,R2)
diff --git a/3886/CH2/EX2.13/2_13.txt b/3886/CH2/EX2.13/2_13.txt new file mode 100644 index 000000000..28ecd955d --- /dev/null +++ b/3886/CH2/EX2.13/2_13.txt @@ -0,0 +1,13 @@ + --> //To determine force in the bar and floor reaction
+--> //Refer Fig. 2.17(b)
+--> //Equilibrium equation gives
+--> S=(7*cosd(45)-5)/cosd(30) //kN
+ S =
+
+ -0.0580266
+--> R=10+7*sind(45)-S*sind(30) //kN
+ R =
+
+ 14.978761
+--> printf("Tensile force in the bar has magnitude %.3f kN and Reaction from floor is R=%.3f kN",-S,R)
+Tensile force in the bar has magnitude 0.058 kN and Reaction from floor is R=14.979 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.13/Ex2_13.sce b/3886/CH2/EX2.13/Ex2_13.sce new file mode 100644 index 000000000..76eb251fe --- /dev/null +++ b/3886/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,6 @@ +//To determine force in the bar and floor reaction
+//Refer Fig. 2.17(b)
+//Equilibrium equation gives
+S=(7*cosd(45)-5)/cosd(30) //kN
+R=10+7*sind(45)-S*sind(30) //kN
+printf("Tensile force in the bar has magnitude %.3f kN and Reaction from floor is R=%.3f kN",-S,R)
diff --git a/3886/CH2/EX2.14/2_14.txt b/3886/CH2/EX2.14/2_14.txt new file mode 100644 index 000000000..7a8b93f7d --- /dev/null +++ b/3886/CH2/EX2.14/2_14.txt @@ -0,0 +1,47 @@ + --> // Finding magnitude of F
+--> //When F is applied at point B,refer fig 2.18(a)and(b)
+--> //From triangle AOC
+--> OC=300-150
+ OC =
+
+ 150.
+--> AO=300
+ AO =
+
+ 300.
+--> alpha=acosd(OC/AO) //degree
+ alpha =
+
+ 60.
+--> //from triangle AOB using geometry we get angle OBA=30 degree
+--> //Resolving the forces we get
+--> R=2000/cosd(30) //N
+ R =
+
+ 2309.4011
+--> F=R*sind(30) //N
+ F =
+
+ 1154.7005
+--> printf("Least force through point B is F=%.1f N",F)
+Least force through point B is F=1154.7 N--> //Least force required through the centre of roller
+--> //Assume that F makes an angle theta with the horizontal
+--> //Refer fig. 2.19 (a) and (b)
+--> //Resolving the forces we get
+--> //F*cosd(theta)=R*sind(60)...(1)
+--> //F*sind(theta)+R*cosd(60)=W...(2)
+--> //Solving (1) and (2) we get
+--> //sind(theta)+cotd(60)*cosd(theta)=W/F
+--> //For obtaining maximum value of W/F we differentiate W/F w.r.t. theta and we get
+--> theta=acotd(cotd(60)) //degree
+ theta =
+
+ 60.
+--> //Least value of F is observed when it is at right angle to reaction R
+--> Fmin=2000*sind(60) //N
+ Fmin =
+
+ 1732.0508
+--> printf("\nLeast force through the centre of roller is Fmin=%.0f N",Fmin)
+
+Least force through the centre of roller is Fmin=1732 N
\ No newline at end of file diff --git a/3886/CH2/EX2.14/Ex2_14.sce b/3886/CH2/EX2.14/Ex2_14.sce new file mode 100644 index 000000000..4d9c976e3 --- /dev/null +++ b/3886/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,25 @@ +// Finding magnitude of F
+//When F is applied at point B,refer fig 2.18(a)and(b)
+//From triangle AOC
+OC=300-150
+AO=300
+alpha=acosd(OC/AO) //degree
+//from triangle AOB using geometry we get angle OBA=30 degree
+//Resolving the forces we get
+R=2000/cosd(30) //N
+F=R*sind(30) //N
+printf("Least force through point B is F=%.1f N",F)
+//Least force required through the centre of roller
+//Assume that F makes an angle theta with the horizontal
+//Refer fig. 2.19 (a) and (b)
+//Resolving the forces we get
+//F*cosd(theta)=R*sind(60)...(1)
+//F*sind(theta)+R*cosd(60)=W...(2)
+//Solving (1) and (2) we get
+//sind(theta)+cotd(60)*cosd(theta)=W/F
+//For obtaining maximum value of W/F we differentiate W/F w.r.t. theta and we get
+theta=acotd(cotd(60)) //degree
+//Least value of F is observed when it is at right angle to reaction R
+Fmin=2000*sind(60) //N
+printf("\nLeast force through the centre of roller is Fmin=%.0f N",Fmin)
+
diff --git a/3886/CH2/EX2.15/2_15.txt b/3886/CH2/EX2.15/2_15.txt new file mode 100644 index 000000000..c4e3258fb --- /dev/null +++ b/3886/CH2/EX2.15/2_15.txt @@ -0,0 +1,14 @@ + --> //Determining the forces in bars AB and AC
+--> //Refer fig 2.20(a) and (b)
+--> //Select AB and AC as x and y axes
+--> //Resolving the forces we get
+--> F1=0 //N
+ F1 =
+
+ 0.
+--> F2=40*cosd(30) //N
+ F2 =
+
+ 34.641016
+--> printf("Force in bar AB is F1=%.0f N and force in bar AC is F2=%.1f N",F1,F2)
+Force in bar AB is F1=0 N and force in bar AC is F2=34.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.15/Ex2_15.sce b/3886/CH2/EX2.15/Ex2_15.sce new file mode 100644 index 000000000..45826bb0f --- /dev/null +++ b/3886/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,7 @@ +//Determining the forces in bars AB and AC
+//Refer fig 2.20(a) and (b)
+//Select AB and AC as x and y axes
+//Resolving the forces we get
+F1=0 //N
+F2=40*cosd(30) //N
+printf("Force in bar AB is F1=%.0f N and force in bar AC is F2=%.1f N",F1,F2)
diff --git a/3886/CH2/EX2.16/2_16.txt b/3886/CH2/EX2.16/2_16.txt new file mode 100644 index 000000000..46f93d825 --- /dev/null +++ b/3886/CH2/EX2.16/2_16.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_16.sce', -1)
+
+the various forces are:-
+T1=224.1 N
+T2=183.0 N
+T3=336.6 N
+T4=326.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.16/Ex2_16.sce b/3886/CH2/EX2.16/Ex2_16.sce new file mode 100644 index 000000000..36811d549 --- /dev/null +++ b/3886/CH2/EX2.16/Ex2_16.sce @@ -0,0 +1,11 @@ +//Forces in various segments of cable
+//Refer fig. 2.21 (a) and (b)
+//Apply Lami's theorem at point D
+T1=250*sind(180-60)/sind(60+45) //N
+T2=250*sind(90+45)/sind(60+45) //N
+//Now consider system of forces acting at B
+//Resolving the forces we get
+T3=(T2*cosd(60)+200)/cosd(30) //N
+T4=T3*sind(30)+T2*sind(60) //N
+printf("\nthe various forces are:-\nT1=%.1f N\nT2=%.1f N\nT3=%.1f N\nT4=%.1f N",T1,T2,T3,T4)
+
diff --git a/3886/CH2/EX2.17/2_17.txt b/3886/CH2/EX2.17/2_17.txt new file mode 100644 index 000000000..c48f19d89 --- /dev/null +++ b/3886/CH2/EX2.17/2_17.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_17.sce', -1)
+The load required to be connected at point D is W=2863.6 N
\ No newline at end of file diff --git a/3886/CH2/EX2.17/Ex2_17.sce b/3886/CH2/EX2.17/Ex2_17.sce new file mode 100644 index 000000000..8def98ef6 --- /dev/null +++ b/3886/CH2/EX2.17/Ex2_17.sce @@ -0,0 +1,12 @@ +//Load required at point D
+//Refer fig. 2.22 (a),(b) and (c)
+//Using simple geometry we have
+alpha=acosd(1.3125/1.5) //degree
+Beta=acosd(2-1.3125) //degree
+//Applying Lami's Theorem at point C
+T1=1500*sind(90)/sind(180-alpha) //N
+T2=1500*sind(90+alpha)/sind(180-alpha) //N
+//Applying Lami's Theorem at point B
+T3=T2*sind(90)/sind(90+Beta) //N
+W=T2*sind(180-Beta)/sind(90+Beta) //N
+printf("The load required to be connected at point D is W=%.1f N",W)
diff --git a/3886/CH2/EX2.18/2_18.txt b/3886/CH2/EX2.18/2_18.txt new file mode 100644 index 000000000..45993c3de --- /dev/null +++ b/3886/CH2/EX2.18/2_18.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_18.sce', -1)
+
+The required values are:-
+T1=44.80 kN
+T2=29.24 kN
+T3=25.04 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.18/Ex2_18.sce b/3886/CH2/EX2.18/Ex2_18.sce new file mode 100644 index 000000000..04b2f780d --- /dev/null +++ b/3886/CH2/EX2.18/Ex2_18.sce @@ -0,0 +1,17 @@ +//finding tension and inclination
+//refer fig. 2.23 (a),(b) and (c)
+//Applying Lami's theorem at B
+T1=20*sind(50)/sind(180+30-50) //kN
+T2=20*sind(180-30)/sind(180+30-50) //kN
+//now consider equilibrium of forces at point C we get
+//T3*sind(theta)=22.4...(1)
+//T3*cosd(theta)=11.20...(2)
+//from (1) and (2) we get
+theta=atand(2) //degree
+//then (1) gives
+T3=T2*sind(50)/sind(theta) //kN
+printf("\nThe required values are:-\nT1=%.2f kN\nT2=%.2f kN\nT3=%.2f kN",T1,T2,T3)
+
+
+
+
diff --git a/3886/CH2/EX2.19/2_19.txt b/3886/CH2/EX2.19/2_19.txt new file mode 100644 index 000000000..45ecb131a --- /dev/null +++ b/3886/CH2/EX2.19/2_19.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_19.sce', -1)
+
+The required values are:-
+T1=38.97 kN
+T2=23.85 kN
+T3=22.50 kN
+theta=54.77 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.19/Ex2_19.sce b/3886/CH2/EX2.19/Ex2_19.sce new file mode 100644 index 000000000..609b37097 --- /dev/null +++ b/3886/CH2/EX2.19/Ex2_19.sce @@ -0,0 +1,21 @@ +//determine tension and inclination
+//Refer fig. 2.24 (a),(b) and (c)
+//consider equilibrium at point B,we get
+//T2*sind(theta)=T1*sind(30)...(1)
+//T2*cosd(theta)=T1*sind(30)-20...(2)
+//consider equilibrium at point C,we get
+//T2*sind(theta)=T3*sind(60)...(3)
+//T2*cosd(theta)=-T3*cosd(60)+25...(4)
+//solving (1) and (3) we get
+//T1=T3*sqrt(3)...(5)
+//solving (2) and (4) and substituting (5) we get
+T3=45/2 //kN
+T1=T3*sqrt(3) //kN
+//then (1)/(2) gives
+theta=atand(1.416) //degree
+T2=19.48/sind(theta) //kN
+printf("\nThe required values are:-\nT1=%.2f kN\nT2=%.2f kN\nT3=%.2f kN\ntheta=%.2f degree",T1,T2,T3,theta)
+
+
+
+
diff --git a/3886/CH2/EX2.2/2_2.txt b/3886/CH2/EX2.2/2_2.txt new file mode 100644 index 000000000..52ef601ef --- /dev/null +++ b/3886/CH2/EX2.2/2_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_2.sce', -1)
+Horizontal and vertical components respectively are:-
+ Fx=10.00 kN (towards left)
+ Fy=17.32 kN (Downward)
\ No newline at end of file diff --git a/3886/CH2/EX2.20/2_20.txt b/3886/CH2/EX2.20/2_20.txt new file mode 100644 index 000000000..ea1a888a6 --- /dev/null +++ b/3886/CH2/EX2.20/2_20.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_20.sce', -1)
+
+The reactions are:-
+RA=433.0 N
+RB=250.0 N
+RC=721.7 N
+RD=577.4 N
\ No newline at end of file diff --git a/3886/CH2/EX2.20/Ex2_20.sce b/3886/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..92234f9ba --- /dev/null +++ b/3886/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,15 @@ +//Reactions developed at contacts
+//Refer fig. 2.25(a),(b and (c)
+//consider equilibrium of cylinder 1
+//using conditions of equilibrium we get
+RA=500*cosd(30) //N
+RB=500*sind(30) //N
+//Consider equilibrium of cylinder 2
+//using conditions of equilibrium we get
+RC=(500+250*sind(30))/cosd(30) //N
+RD=RC*sind(30)+250*cosd(30) //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+
+
+
+
diff --git a/3886/CH2/EX2.21/2_21.txt b/3886/CH2/EX2.21/2_21.txt new file mode 100644 index 000000000..d83233754 --- /dev/null +++ b/3886/CH2/EX2.21/2_21.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_21.sce', -1)
+
+The reactions are:-
+RA=1066.7 N
+RB=1333.3 N
+RC=3066.7 N
+RD=2828.4 N
\ No newline at end of file diff --git a/3886/CH2/EX2.21/Ex2_21.sce b/3886/CH2/EX2.21/Ex2_21.sce new file mode 100644 index 000000000..06c4ef72b --- /dev/null +++ b/3886/CH2/EX2.21/Ex2_21.sce @@ -0,0 +1,14 @@ +//reactions developed at contact surfaces
+//Refer fig. 2.26 (a),(b) and (c)
+//using geometry
+theta=acosd(0.8) //degree
+//consider equilibrium of cylinder 1
+//Using equilibrium conditions
+RB=800/sind(theta) //N
+RA=RB*cosd(theta) //N
+//consider equilibrium of cylinder 2
+//Using equilibrium conditions
+RD=((RB*sind(theta))+1200)/cosd(45) //N
+RC=RD*sind(45)+RB*cosd(theta) //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+
diff --git a/3886/CH2/EX2.22/2_22.txt b/3886/CH2/EX2.22/2_22.txt new file mode 100644 index 000000000..23d53e936 --- /dev/null +++ b/3886/CH2/EX2.22/2_22.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_22.sce', -1)
+
+The reactions are:-
+RA=200.0 N
+RB=600.0 N
+RC=200.0 N
+RD=632.5 N
\ No newline at end of file diff --git a/3886/CH2/EX2.22/Ex2_22.sce b/3886/CH2/EX2.22/Ex2_22.sce new file mode 100644 index 000000000..76e7694fb --- /dev/null +++ b/3886/CH2/EX2.22/Ex2_22.sce @@ -0,0 +1,11 @@ +//determine the reactions developed at contact points
+//refer fig. 2.27 (a),(b) and (c)
+//considering the equilibrium conditions of cylinders we have
+RB=600 //N
+alpha=atand(450/150) //degree
+RD=RB/sind(alpha) //N
+RC=RD*cosd(alpha) //N
+RA=RC //N
+printf("\nThe reactions are:-\nRA=%.1f N\nRB=%.1f N\nRC=%.1f N\nRD=%.1f N",RA,RB,RC,RD)
+//The answers vary due to round off error
+
diff --git a/3886/CH2/EX2.23/2_23.txt b/3886/CH2/EX2.23/2_23.txt new file mode 100644 index 000000000..a07c5897b --- /dev/null +++ b/3886/CH2/EX2.23/2_23.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_23.sce', -1)
+P=1071.8 N is required to hold the system in given position.
\ No newline at end of file diff --git a/3886/CH2/EX2.23/Ex2_23.sce b/3886/CH2/EX2.23/Ex2_23.sce new file mode 100644 index 000000000..e3bb15dd9 --- /dev/null +++ b/3886/CH2/EX2.23/Ex2_23.sce @@ -0,0 +1,8 @@ +//Force P required to hold the system
+//Refer to fig. 2.28 (a),(b)&(c)
+//Applying Lami's Theorem at A we get
+C=4000*sind(180-60)/sind(60+90-15) //N
+//Applying equilibrium conditions B
+P=(-2000*cosd(45)+C*cosd(60))/cosd(15) //N
+printf("P=%.1f N is required to hold the system in given position.",P)
+
\ No newline at end of file diff --git a/3886/CH2/EX2.3/2_3.txt b/3886/CH2/EX2.3/2_3.txt new file mode 100644 index 000000000..e8cf16c3a --- /dev/null +++ b/3886/CH2/EX2.3/2_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_3.sce', -1)
+The normal and parallel components respectively are :-
+ Wn=9.40 kN
+ Wp=3.42 kN
\ No newline at end of file diff --git a/3886/CH2/EX2.3/Ex2_3.sce b/3886/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..f25a1debc --- /dev/null +++ b/3886/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,6 @@ +//Components of block normal to and parallel to inclined plane
+//Let Wn be the normal component and Wp be the parallel component
+//Refer fig. 2.5(b),triangle ABC
+Wn=10*cosd(20) //kN
+Wp=10*sind(20) //kN
+printf("The normal and parallel components respectively are :-\n Wn=%.2f kN\n Wp=%.2f kN",Wn,Wp)
diff --git a/3886/CH2/EX2.4/2_4.txt b/3886/CH2/EX2.4/2_4.txt new file mode 100644 index 000000000..36d030bf0 --- /dev/null +++ b/3886/CH2/EX2.4/2_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_4.sce', -1)
+
+The resultant is R=161.5 N
+The inclination of resultant w.r.t. positive x-axis is alpha=18.81 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.4/Ex2_4.sce b/3886/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..76affa39e --- /dev/null +++ b/3886/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,8 @@ +//Resultant of three forces that are acting on a hook
+//Resolving all forces along x and y axis gives
+Fx=70*cosd(50)+80*cosd(25)+50*cosd(45) //N
+Fy=70*sind(50)+80*sind(25)-50*sind(45) //N
+R=sqrt(Fx^2+Fy^2) //N
+alpha=atand(Fy/Fx) //degree
+printf("\nThe resultant is R=%.1f N \nThe inclination of resultant w.r.t. positive x-axis is alpha=%.2f degree",R,alpha)
+//The answers vary due to round off error
diff --git a/3886/CH2/EX2.5/2_5.txt b/3886/CH2/EX2.5/2_5.txt new file mode 100644 index 000000000..c8117022d --- /dev/null +++ b/3886/CH2/EX2.5/2_5.txt @@ -0,0 +1,4 @@ + The resultant of given forces :-
+R=160.2 N.
+Inclination of resultant w.r.t X-axis :-
+alpha=24.1 degree.
\ No newline at end of file diff --git a/3886/CH2/EX2.5/Ex2_5.sce b/3886/CH2/EX2.5/Ex2_5.sce new file mode 100644 index 000000000..c7c6c6239 --- /dev/null +++ b/3886/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,17 @@ +//Determining the resultant
+clc
+//Given-
+//inclination of 200N force with x axis calculated by using slope of 200N force as shown in fig.2.8
+theta1=atand(1/2) //degree
+//inclination of 120N force with x axis calculated by using slope of 120N force as shown in fig.2.8
+theta2=atand(4/3) //degree
+//summation of forces in X direction
+Fx=200*cosd(26.565)-120*cosd(53.13)-50*cosd(60)+100*sind(40) //N
+//summation of forces in Y direction
+Fy=200*sind(26.565)+120*sind(53.13)-50*sind(60)-100*cosd(40) //N
+//Resultant
+R=sqrt((Fx)^2+(Fy)^2) //N
+//inclination of resultant w.r.t X axis
+alpha=atand(65.5/146.2) //degree
+printf("The resultant of given forces :-\nR=%.1f N.\n",R)
+printf("Inclination of resultant w.r.t X-axis :-\nalpha=%.1f degree.",alpha)
diff --git a/3886/CH2/EX2.6/2_6.txt b/3886/CH2/EX2.6/2_6.txt new file mode 100644 index 000000000..748e11388 --- /dev/null +++ b/3886/CH2/EX2.6/2_6.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_6.sce', -1)
+The resultant has magnitude R=234 N directed up the plane
\ No newline at end of file diff --git a/3886/CH2/EX2.6/Ex2_6.sce b/3886/CH2/EX2.6/Ex2_6.sce new file mode 100644 index 000000000..f47c7d20f --- /dev/null +++ b/3886/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,12 @@ +//Determine Resultant Force
+//From given data
+T=1200 //N
+F=100 //N
+N=500 //N
+W=1000 //N
+theta=60 //degree
+//Taking co-ordinate system parallel and perpendicular to plane as x and y axis and resolving the forces
+Fx=T-F-W*sind(theta) //N
+Fy=N-W*cosd(theta) //N
+R=sqrt(Fx^2+Fy^2)
+printf("The resultant has magnitude R=%.0f N directed up the plane",R)
diff --git a/3886/CH2/EX2.7/2_7.txt b/3886/CH2/EX2.7/2_7.txt new file mode 100644 index 000000000..52fc22652 --- /dev/null +++ b/3886/CH2/EX2.7/2_7.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_7.sce', -1)
+The third force is F=467.2 N and makes an angle of theta=61.08 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.7/Ex2_7.sce b/3886/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..71b465ef7 --- /dev/null +++ b/3886/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,10 @@ +//Finding the third force F
+//Assume that the third force F makes an angle theta with x-axis
+//Resolving the forces we get
+//F*cosd(theta)=-225.9...(1) //N
+//F*sind(theta)=-408.9...(2) //N
+//Then (2)/(1) gives
+theta=atand(-408.9/-225.9) //degree
+F=sqrt(225.9^2+408.9^2) //N
+printf("The third force is F=%.1f N and makes an angle of theta=%.2f degree",F,theta)
+
diff --git a/3886/CH2/EX2.8/2_8.txt b/3886/CH2/EX2.8/2_8.txt new file mode 100644 index 000000000..ee13f3330 --- /dev/null +++ b/3886/CH2/EX2.8/2_8.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_8.sce', -1)
+Required value of theta=6.31 degree
\ No newline at end of file diff --git a/3886/CH2/EX2.8/Ex2_8.sce b/3886/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..398e033aa --- /dev/null +++ b/3886/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,6 @@ +//Determining the value of theta
+//x and y axes are selected as shown in fig. 2.11
+//As the resultant is directed along x-axis,component of resultant in y-direction is zero
+//Fy=0 gives
+theta=(asind(0.833/(2*cosd(20))))-20 //degree
+printf("Required value of theta=%.2f degree",theta)
diff --git a/3886/CH2/EX2.9/2_9.txt b/3886/CH2/EX2.9/2_9.txt new file mode 100644 index 000000000..950590c99 --- /dev/null +++ b/3886/CH2/EX2.9/2_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\2. Resultant and equilibrium of system of coplanar concurrent forces\Ex2_9.sce', -1)
+
+The required values are:-
+T=103.5 N
+R=26.8 N
\ No newline at end of file diff --git a/3886/CH2/EX2.9/Ex2_9.sce b/3886/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..6a5b0b890 --- /dev/null +++ b/3886/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,5 @@ +//Finding T and R
+//applying Lami's Theorem we get
+T=(100*sind(90))/sind(90+15) //N
+R=(100*sind(180-15))/sind(90+15) //N
+printf("\nThe required values are:-\nT=%.1f N \nR=%.1f N",T,R)
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