Static EQ/Center of Mass problem

[Ben is Dumb]

Homework Statement

A 65.0-kg person stands on a horizontal board. The board is 10.0-m long and is NOT perfectly uniform. The person stands 1/4 of the way from the left end of the board. The board is supported at each end by a scale. The scale on the left reads 753 N and the scale on the right reads 619 N.

What is the mass of the board?
How far from the left end is the board's center of mass?

My digital drawing skills are just as solid as my math skills.

Homework Equations

None/unsure

The Attempt at a Solution

I was able to get the mass of the board (75kg) but unable to calculate the center of mass. It seems like integration would do the trick, but my class has been a bit skimpy on the calculus and I'm unsure how to set it up. Otherwise balancing the torques makes sense, but I can't get rid of the person's influence on the scales. I apologize if this isn't much of an attempt at a solution, but I've tried to look at this thing a bunch of different ways and can't seem to get very far.

Thanks for the help!

Ben

 

[PhanthomJay]

You've found the mass of the board, so now find its weight. You should be familiar with Newton's first law of rotational equilibrium: The sum of the torques of all forces about any point must equal zero. So try summing torques of all forces about the left end, to solve for the unknown distance of the center of the boards mass from the left end.

 

[Ben is Dumb]

Alright, here's what I did.

(CCW is positive)
Torque= -75kg(9.8m/s^2) X(COM) + 2.5m(65kg)(9.8m/s^2) = 0

I actually got a reasonable answer (2.17m), just not the right one (6.26m). I must still be setting something up the wrong way. Thanks for the help Jay.

 

[PhanthomJay]

The boards weight and the person's weight both produce clockwise torques about the left end, so why did you choose plus for the person's weight? Also, what about the torque from the right end reaction?

 

[Ben is Dumb]

I talked to my prof, fixed the signs and put the 619N(L) on the other side and got the correct answer. Thanks again.