A 10-m long plank (of negligible mass)is supported at each end by vertical cables.A person of unknown weight sits on the plank between the cables.The tension in the left cable is 300 N,and in the right cable it is 200 N.How far is the person sitting from the left cable?
A)2.0 m
B)3.0 m
C)4.0 m
D)6.0 m
E)7.0 m

Question

Quizplus · Accepted Answer

When the plank is in equilibrium, the sum of the upward forces acting on it must equal the sum of the downward forces.
Let T1 be the tension in the left cable, T2 be the tension in the right cable, and W be the weight of the person. 
The weight of the plank itself is negligible and can be ignored. 
For the plank to be in equilibrium:
T1 + T2 = W
We know that T1 = 300N and T2 = 200N, so:
300N + 200N = W
W = 500N
Now, we can find the distance, x, that the person is sitting from the left cable by considering the torque equation about the left cable:
T1(x) = T2(10m - x)
Substituting for T1 and T2 and solving for x:
300N(x) = 200N(10m - x)
300x = 2000 - 200x
500x = 2000
x = 4m
Therefore, the person is sitting 4m from the left cable. 
So, the answer is C) 4.0 m.

A 10-M Long Plank (Of Negligible Mass)is Supported at Each