# Equilibrium non-uniform Box

1. Apr 25, 2017

### alex91alex91alex

1. The problem statement, all variables and given/known data
A non-uniform box sits on two scales as seen below. Scale A reads 10.2N, and scale B reads 19.4N.

Question: How far from the left end of the box is the center of mass located?

2. Relevant equations
τccw=τcw
τ=f*d(from pivot point)

3. The attempt at a solution
If I choose the pivot point to be the left end side of the box I get
10.2N*d1=19.4N*d2
I know that at the pivot point the box should be in equilibrium but how am I suppose to solve the equation with 2 unknowns? Do I choose a different pivot point, or am I using the wrong formula?

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2. Apr 25, 2017

### kuruman

You need to define x as the unknown distance of the CM from the left end of the box. There are three forces generating torques in this problem, gravity, scale A and scale B.

3. Apr 25, 2017

### Staff: Mentor

Can you say what the total weight of the box is?

4. Apr 25, 2017

### alex91alex91alex

The total weight would be (10.2N+19.4N)/g=3.02kg, right?

5. Apr 25, 2017

### alex91alex91alex

Hmm, would the mg of the box be CW or CCW?

6. Apr 25, 2017

### Staff: Mentor

That would be the total mass. Just sum the forces for the total weight in Newtons. So w = 10.2N+19.4N.
It's neither until you pick a center of rotation

Draw your a FBD with mg located at some (unknown) distance x from your chosen center of rotation.

7. Apr 25, 2017

### alex91alex91alex

Ok, so lets says the center of rotation is the left end point.

10.2N(0.02m)+19.4N(0.3m-0.074m)=(10.2N+19.4N)x
10.2N(0.02m)+19.4N(0.226m)=29.6N*x
x=0.155m

is that right?

8. Apr 25, 2017

### kuruman

That is right.