# Angular momentum, and friction

## Homework Statement

There is a box with a width of 2a, a height of 2b, and a mass of m. There is a rope pulling the box from the one of the upper corners.
(a) assume the floor is frictionless. What is the maximum force that can be applied without causing the box to tip over. hint: use the center of mass

(b) Repeat part (a) but now let the coeffiecent of friction be mu.

(c) What happens to your anwer to part (b) when the box is sufficiently tall? How do you interpret this?

## Homework Equations

Torque= Force * distance from center of mass
Tension = F(rope) + Force(friction)
Statics: the total torque must equal zero.
Friction=Mu*FN

## The Attempt at a Solution

Part (a) was easy.
max torque = 0 = Torque mass + torque rope
0 = F * sin 90 +F(rope)*sin angle

using the center of mass sin ∅ = a/(b^2+a^2)^0.5
using the center of mass cos ∅ = b/(b^2+a^2)^0.5
using the center of mass tan ∅ = a/b

F comp-y
0=F*cos ∅-mg; rope vs block
F= mg/cos∅

F comp-x
Tension(rope) = F*sin ∅
Tension = mg/cos∅ *sin∅
"Tension = mg (a/b)"

Part (b) has me a little stumped as I am not sure about the amount of torque generated by the friction
The torque is close to the same equation but the friction is going to have the same rotation as the rope. So Total torque is = to Torque mass + torque rope+torque friction. I tried the F*Mu didn't work so, that is where I became stuck.

Simon Bridge
Homework Helper
You seem to be on the right track...
Draw the friction arrow on the diagram to work out the moment arm etc.
Where does the friction force act?

Between the block, and the table. The total tension on the rope is equal to force rope + force friction.
Now the torque of the rope, and the torque of the friction will have the same direction of rotation. So for the block to not tip the two torque can not be greater than the torque of gravity. The problem is that I am not sure how that part of the torque equation should work....Since the distance from the center of mass isn't uniform

Simon Bridge
Homework Helper
Did you draw the picture?

The friction force acts between the block and the table - sure - but: where between the block and the table? This is the bit you are stuck on right?

Reason it out - friction is calculated from the normal force, where does that act?
If the block started to tilt though, friction would act from the forward corner ... is this valid if it doesn't tilt?

Does the applied Force change the Normal force?

--- aside: ---------------
Trouble is, this is homework so I'm trying to give you a guide for your thoughts without actually giving stuff away.

At some point, you'll get a grade and an official answer. If you are still confused about how that was obtained, show us the model answer and ask more questions. But until that time, getting a quick answer will be a bit like pulling teeth :/ But give it a go - if you get most of the way I'll give you a leg-up the last bit.

Yes, I have a picture.
I know the friction is occuring between the surface of the table, and the bottom surface of the block with a length of 2a. I know where but the how to write it symbolically, and the ratio is the sticky point.

__________________rope
[ c ]
[ e ]
[ n ]
[ t ]
__[___]_________table

if it tilts then the only friction is at the point, not the whole surface area. But that does give me an idea.

Simon Bridge
Homework Helper
Yep - friction does not depend on the area of contact.
Let me know how you get on.

Part of the issue I realized after working with two tutors, and another physics professor was my lack of understanding of how torque can be calculated as my professor just wrote a one sentence description of it.....

0=Trope+Tgravity+Tfriction+Ttable
The force of the rope remained as F, and noted as such so that I wouldn't convert it to mg
0=b*F+mg*sin 90 + b*FN*mu -a*FN
mg*sin 90 is = to zero
0=b*F+b*FN*mu-a*FN
b*F+b*FM*mu=a*FN
b*F+b*mg*mu=a*mg
b(F+mg*mu)=a*mg
F+mg*mu=(a/b)mg
F(rope)=(a/b)mg-mg*mu

Now my professor's answer looks like this:
(a) Let's use subscripts "r" for the rope, "g" for gravity, and "n" for the floor's normal force. In the case where the block topples, it rotates and its center of mass rises up and over. In the case where it slides without tipping, neither of these is supposed to happen, so vertical forces must cancel, and the torques must cancel. Cancellation of the vertical forces fives Fn=mg. The torques are most easily calculated using t=r*F, with the axis taken to be at the center of the box. Note that in the condition where the box is not tipping, the normal force is actually being distributed over the whole bottom surface of the block, so it doesn't actually have a single well-defined value of r; however, as soon as the tipping starts, the only point of contact is at r=a. The torques are therefore t=bF, and T=amg. If the box is just about to tip, i.e., the force from the rope has its maximum value, then these torques cancel, and F=(a/b)mg. The dependence on a/b makes sense, because a squat box with a large value of a would be more stable, allowing F to be greater.(What we've actually calculated here is the value of F for which the box would just barely start to tip. Once the box started to tip, we could ask ourselves whether it might then manage to right itself again. The answer is no, because once it starts to tip, both of the r values start to decrease, but their ratio is still (a/b)
(b)The force of friction is F=mu*F=mu*mg. (In the case of static friction, this is the maximum force, which is what we're interested in because we want to see if we can unstick the box and get it to slide without tipping. In the case of kinectic friction, this is simply the value of the frictional force) We now have a torque from friction, which is in the same clockwise direction as the torque from the rope. Cancellation of the torques fives b*F +b*mu*mg=a*mg, yielding F=(a/b-mu)mg.
(c) the answer to part b, unlike the one from part (a), could come out negative under certain circumstances. the negative solution would be unphysical, and the interpretation would be that it would be impossible to accelerate the box without tipping it over. This happens for b >a/mu. That makes sense. A big value of b means a tall box, which would be less stable. In the limit mu -> 0, we get b -> infinity, meaning that there is no maximum value of b, recovering the behavior of the frictionless case.