Tension on two chains holding a board

[Physsics]

Homework Statement

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighting 500 N is sitting on the board. The tension in the right chain is:

a) 250 N
b) 375 N
c) 500 N
d) 625 N
e) 875 N

How far is the person from the left end of the board?
a) .4 m
b) 1.5 m
c) 2 m
d) 2.5 m
e) 3 m

I really have no idea how to do this one; wouldn't you have to know how far the person is from one of the sides first to know what the tension is on one side?

This is my first post here, so I hope I didn't do something wrong :)

 

[rock.freak667]

Start by drawing a free body diagram and then use the fact that the board is in equilibrium.

 

[Physsics]

So the body is 500 N downwards, while the 250 N tension, the unknown tension of the second chain, and the 125 N board all oppose it, and the fact that it's in equilibrium means that the sum of all the net forces is 0, but what after that?

 

[rock.freak667]

So the body is 500 N downwards, while the 250 N tension, the unknown tension of the second chain, and the 125 N board all oppose it, and the fact that it's in equilibrium means that the sum of all the net forces is 0, but what after that?

How did you get the tension in the left chain?

 

[Physsics]

I didn't, I just said that one of the chains (presumably the right chain) has a tension of 250 N.

 

[rock.freak667]

I didn't, I just said that one of the chains (presumably the right chain) has a tension of 250 N.

Well you'd have the two weights downwards and the two tensions upwards. Then you should be able to use ∑F_y=0 to get the tensions as they should be the same.