# Moment and torque physics problem

1. Sep 8, 2009

### stpmmaths

1. The problem statement, all variables and given/known data

2. Relevant equations

Equilibrium- Moment/Torque

3. The attempt at a solution

The answer given is x = L/2, y = L/4, z = L/6.
What should I do to find these answer? (Those answers may or may not be correct)

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2. Sep 8, 2009

### kuruman

Re: Moment/Torque

The statement of the problem did not come through. You may have to type it in. Also you have to make a better effort. Equilibrium- Moment/Torque is not an equation. If you don't know how to get the answer, at least tell us what you know about the problem, what it is about and what you think might be relevant in your effort to get to the answer.

3. Sep 8, 2009

### kuruman

Re: Moment/Torque

Ah! The picture came through.

Suppose that you had only one block resting at the edge of a table so that x meters hang over the edge. Do you understand why x = L/2? Why not larger? Why not smaller? Answer this question and you will understand why the top block was placed as shown. Then worry about the other blocks.

4. Sep 9, 2009

### stpmmaths

Re: Moment/Torque

Actually my main problem is I don't know why x = L/2. Can you show me the force acting on the top block that cause it to be in equilibrium?
Thanks.

5. Sep 9, 2009

### kuruman

Re: Moment/Torque

The forces acting on the top block are gravity and the normal force exerted by the body supporting it. Where, i.e. at what point, does the force of gravity act on the top block?

6. Sep 9, 2009

### saunderson

Re: Moment/Torque

Hi stpmmaths,

i have another approach. Think about the http://en.wikipedia.org/wiki/Center_of_mass" [Broken] of each block! That means, that each block can be imagined as a point mass, at a certain Point.

So the x-component (see sketch below) of the center of mass of a bunch of blocks must be on the edge of the block below (acutally an infinitesimal piece forward the edge, to be in stable equilibrium, otherwise the "point mass" falls down!)

Cause my english is not so well, i've made a sketch:

So considered from the origin of the coordinate system i have chosen, i declare the x component of the center of mass of the blocks with
• Block 1 : $$r^{^{(cm)}}_{x,1}$$
• Block 2 : $$r^{^{(cm)}}_{x,2}$$
• Block 3 : $$r^{^{(cm)}}_{x,3}$$

with $$R_x = \frac{1}{\sum_i m_i} \, \sum_i m_i \cdot r^{^{(cm)}}_{i,x}$$ it yields

$$R_x = \underbrace{L}_{\mbox{edge of the lowest block}} = \frac{1}{3\cdot m_{block}} \cdot m_{block} \cdot \left( \underbrace{\left(z+\frac{L}{2}\right)}_{ \mbox{center of mass B3} } + \underbrace{\left(y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B2} }+ \underbrace{\left(x+y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B1} } \right) = \frac{L}{2} + \frac{x}{3} + \frac{2y}{3} + z$$

if you put the answers for x,y,z in the above equation it results L, like i've promised ;)

But my equation above has infinity results (Cause my equation treats the 3 blocks as one continuum....) , so you must think about what constraints are on x and on y!

TIP: work top down through the blocks

with best regards and i hope i could help

Last edited by a moderator: May 4, 2017
7. Sep 11, 2009

### stpmmaths

Re: Moment/Torque

thanks