Assignment #1
1. The two structural members, one of which is in tension and the other in compression, exert the indicated
forces on joint O. Convert the two forces into a single resultant R. Use both vector sum and parallelogram
methods. (CLO1-C3)
Solution:
 For Parallelogram method of vector
addition
Scale 1cm=1kN
From scale
R=8.9 cm= 8.9kN
θ=-4o
Angle along +x-axis =180-4=176o
From Rectangular components of vector addition.
F1x   F1 cos 60o  4 cos 60o  4  0.5  2kN
F1 y   F1 sin 60o  4sin 60o  4  0.86  3.46kN
F2 x   F2 cos 30o  8cos 30o  8  0.866  6.928kN
F2 y  F2 sin 30o  8sin 30o  8  0.5  4kN
Fx  F1x  F2 x  2  6.928  8.928kN
Fy   F1 y  F2 y  3.46  4  0.54kN
R  Fx2  Fy2 
Fy
 8.928   0.54
2
2
 79.71  0.29  80  8.94kN
0.54
 tan 1  0.06   3.43o  4o
Fx
8.928
Angle along +x-axis =180-4=176o
  tan 1
 tan 1
2. Determine the resultant R of the two forces shown and angle θ which R makes with x-axis by (a) applying
the parallelogram rule for vector addition and (b) summing scalar components. (CLO1-C3)
Solution:
V1  400 N , 1  270o
V1x  400 cos 270o  400  0  0 N
V1 y  400sin 270o  400  1sin 60o  400 N
V1  600 N , 1  330o
V1x  600 cos 330o  600  0.866  519.6 N
V1 y  600sin 330o  600  0.5  300 N
Vx  V1x  V2 x  0 N  519.62 N  519.62 N
Vy  V1 y  V2 y  400   300   700 N
V  Vx2  Vy2 
520   700
2
2
 270400  490000  760400  872 N
V  872 N
per :
519.62
  tan 1
 tan 1
 tan 1  0.7423  36.6o
Base
700
  36.6o
3. As a trailer is towed in the forward direction, the force F = 500 N is applied as shown to the ball of the
trailer hitch. Determine the moment of this force about point O using (a) Varignon’s theorem and (b) vector
approach. (CLO1-C3)
Solution:
F=500N
Vx  500 cos 30o  500  0.866  433 N
Vy  500sin 30o  500  0.5  250 N
Now, the moment of force will be:
 M o  433 275  250  38  119075  9500  128575N  m
M
o
 128575N  m
4. A force F of magnitude 60 N is applied to the gear. Determine the moment of F about point O using (a)
Varignon’s theorem and (b) vector approach. (CLO1-C3)
Solution:
F = 60 N
r = 100mm =100/1000=0.1m
F  Fx  F cos 20  60  0.9396  56.38N
M o  F  r
M o  56.38  0.1  5.64 N  m
M o  5.64 N  m