Question

A horizontal uniform board of weight 125N and length 4 m is supported by vertical chains at each end. A person weighing 500N sitting on the board .Thr tension in the right chain is 250N. How far from left of the end person is sitting

Answer 1

Let's apply an equation of equilibrium to the situation: The sum of the moments about the left end of the board must equal 0.

We have three moments to add. Positive force values indicate upward direction and negative values indicate downward direction. All distances given below are measured to the right side of the left end of the board:

1. The weight of the board, -125N, located at 2m (center of the board due to its uniform density)
2. The tension in the right chain, +250N, located at 4m
3. The weight of the person, -500N, located at a distance "x"

The sum of the moments must equal 0 and is given by:

ΣFx = 0

F is the magnitude of force, x = distance from the left end of the board

Plug in all of the force and distance values and solve for x:

ΣFx = 250(4) - 125(2) - 500x = 0

500x = 750

x = 1.5m