Solve Chain Torque Problem: Find Distance from Left Chain

[Nivlac2425]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Well I've calculated that the left chain has a tension of 375N as to be able to hold the person-board system in equilibrium. But now they want to know how far the person is from the left chain. I know this requires the balancing of torque, but I'm not sure how.
Thanks for helping out!

 

[rl.bhat]

Now take the torque about the left.
Consider the man sitting at a distance x from the left end.
Equate clockwise moment and counterclockwise moment to find x.

 

[nlightner]

I would approach this problem by first identifying the relevant equations and principles involved. In this case, the problem can be solved using the principle of torque, which states that for an object to be in rotational equilibrium, the sum of the torques acting on it must be equal to zero.

To find the distance from the left chain, we can use the equation:

τ = Fd

Where τ is the torque, F is the force applied, and d is the distance from the pivot point (in this case, the left chain).

We know that the person's weight (500N) and the board's weight (125N) are acting downwards, creating a torque in the clockwise direction. The tension in the right chain (250N) is acting upwards, creating a torque in the counterclockwise direction. To maintain equilibrium, these torques must cancel each other out.

Therefore, we can set up the following equation:

(500N + 125N)(d) = (250N)(4m)

Solving for d, we get:

d = 4m(250N)/(500N + 125N)

d = 1.6m

Therefore, the person is sitting 1.6m from the left chain.