Equilibrium of Non-Uniform Box on Two Scales

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A non-uniform box is balanced on two scales, with readings of 10.2N and 19.4N, prompting a calculation of the center of mass distance from the left end. The equilibrium condition is established using torque equations, where the torques from the two scales must balance. The total weight of the box is calculated as the sum of the scale readings, leading to a mass of approximately 3.02kg. By selecting the left end as the pivot point, the distances for the torques are defined, resulting in the center of mass being located 0.155m from the left end. The calculations confirm the accuracy of this distance.
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Homework Statement


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?

Homework Equations


τccw=τcw
τ=f*d(from pivot point)

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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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.
 
Can you say what the total weight of the box is?
 
The total weight would be (10.2N+19.4N)/g=3.02kg, right?
 
Hmm, would the mg of the box be CW or CCW?
 
alex91alex91alex said:
The total weight would be (10.2N+19.4N)/g=3.02kg, right?
That would be the total mass. Just sum the forces for the total weight in Newtons. So w = 10.2N+19.4N.
alex91alex91alex said:
Hmm, would the mg of the box be CW or CCW?
It's neither until you pick a center of rotation :smile:

Draw your a FBD with mg located at some (unknown) distance x from your chosen center of rotation.
 
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Ok, so let's 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?
 
That is right.
 
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