Equilibrium of Non-Uniform Box on Two Scales

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Homework Help Overview

The problem involves determining the location of the center of mass of a non-uniform box that is resting on two scales, with given readings from each scale. The context is centered around the principles of equilibrium and torque in physics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of torque equations to find the center of mass, with one participant questioning the choice of pivot point and the implications of having two unknowns. Others suggest defining the distance of the center of mass and consider the forces acting on the system.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and providing guidance on defining variables and setting up equations. There is no explicit consensus, but some productive directions have been suggested regarding the setup of the equations and the role of forces.

Contextual Notes

Participants are navigating the complexities of torque calculations and the implications of choosing different pivot points. There is a focus on understanding the total weight of the box and how it relates to the forces measured by the scales.

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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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