d) A string ABCD attached to fixed points A and D has two equal weights 100 N attached to it at B and C. The weights rest with the portion AB and CD inclined at an angle as shown in figure. Find the tension in the portions AB, BC and CD of the string, if the inclination of the portion BC with the vertical is 120°.
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Since the system is in equilibrium;
The moment of forces about point A should be zero.
Thus;
10g×0.5+15g×1=Tsin53
o
×1
50+150=0.8T
T=250 N
At point A, the hinge force is working which can be resolved as R
y
& R
x
R
x
=Tcos 53
o
R
x
=250×0.6=150 N
R
y
=Tsin53
o
R
y
=250×0.8=200 N
R=
(R
x
)
2
+(R
y
)
2
R=
150
2
+200
2
R=250 N
Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N
solution
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