# Torque with 2D center of mass

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1. May 2, 2017

### WilliamDaFoe

1. The problem statement, all variables and given/known data
A cube of mass M rests tilted against the wall as shown (see below). There is no friction between the wall and the cube, but the friction between the cube and the floor is just sufficient to keep the cube from slipping. When $0\lt\theta\lt 45^\circ$ find the minimum coefficient of friction $\mu_{min}$ as a function of $\theta$

2. Relevant equations

So, we have two points of contant with the wall. The point of contant between the vertical wall and the cube results in an unknown force $N_1$. You then have three forces on the point of contant between the cube and the 'ground'. $N_2$ is the standard normal force. $\mu N_2$ would be a force parallel to the ground going towards the inner wall to oppose any action to move the cube and $F_2$ is the force parallel to the ground going away from the wall. From here force resolution gets us:

$$\sum F_x = N_2 -\mu N_1 = 0$$
$$\sum F_y = N_1 - mg(?)$$
$$\sum M_b = ??$$

3. The attempt at a solution

So for the $F_y$ force my thinking is that because all of the weight of the cube will be pressing down at that one point we can get the $mg$ term. If not it will need to have a value of $\frac{mgl}{\sqrt{2}}cos(\theta+45)$.

For the moment term I made an assumption that we could consider this box 1d and have the mass along the line so we get $M_b=0=\frac{mgl}{2}cos(\theta)-lN_1 sin(\theta)$. This did not work.

I am thinking I need the CM to be at the center of the cube, in which case I have the mg term as mentioned but I am not sure how to find the $M_b$ term in this case.

This problem comes from the problem set for feynman and in the lecture for this he mentions classical field theory. I was also wondering if this problem could be solved using CFT and if it would be a worthwhile exercise to do so. Also, is it possible to solve this using virutal work? And how would I go about doing that? Looking for as many ways to solve this as possible to make sure I can do this later.

Last edited: May 2, 2017
2. May 2, 2017

### haruspex

That's a force the cube exerts on the ground. In analysing the cube, only consider forces acting on the cube.

For the LaTeX, either use doubled \$ signs (which will put the equation on a line by itself) or doubled # signs.

3. May 2, 2017

### WilliamDaFoe

Thank you, fixed.

4. May 3, 2017

### haruspex

Ok, you altered the equation, but you still have F2 in the text, so it reads a bit strangely.

In your force balance equations, you seem to have crossed over N1 and N2 compared with how you defined them.