Find Friction for Cube on Rough Floor

[e(ho0n3]

Need some feedback on my attack strategy to the following:

A cube of side L rests on a rough floor. It is subject to a steady horizontal pull, F, exerted a distance h above the floor. As F is increased, the block will either begin to slide, or begin to tip over. (a) What must be the coefficient of static friction so that the block begins to slide rather than tip? (b) What must be the coefficient of static friciton so that the block begins to tip?

For (a): Let f be the force of friction. Is it sufficient to say that if the block begins to slide, then F - f > 0? Would it be wiser to calculate the torque about the tipping edge and set it equal to 0?

For (b): I'm not exactly sure what to do here. I drew a free-body diagram of the situation, but I don't know how to place F (is it still acting horizontally?). I guess I should consider the instant the block begins to tip (so that F is still acting horizontally). If this is the case, should I calculate the torque about the tipping edge again (with the difference that the normal is now acting entirely from the tipping edge, unlike in (a)).

 

[arildno]

In (a):
Here, we must require that the torque of the weight about the tipping corner is greater in absolute value than the torque from F about the same corner.
(In other words, there exist a a non-zero torque from the normal force to complete the balance (since the force of friction works parallell to its direction from the corner, it doesn't provide a torque)

In addition, when the cube starts slipping, F=f (afterwards, f turns into kinetic friction, i.e,f<F)

Hope this helped..

 

[Zorodius]

I'd just like to point out that this is a bit of a nuanced problem, since, if I'm correct, the most efficient way to tip such a box is actually to push upward at a 45 degree angle and gradually reduce that angle as you tip the box. This way, you have the greatest possible moment arm for your torque at all times. A perfectly horizontal push will initially have a moment arm equal to the height of the box, but as the box turns, that will be increased to the diagonal of the cube.

I don't think it sounds like any of that is supposed to be considered in the answer, though.

 

[e(ho0n3]

In addition, when the cube starts slipping, F=f (afterwards, f turns into kinetic friction, i.e,f<F)

Right. I forgot this friction is static. So then I don't need to fiddle with the torque to find the answer to (a), I can just use F = f?

 

[arildno]

You need the inequality of the torques I mentioned..
This gives you one equation and one inequality; the inequality may be solved as a bound for the coefficient of static friction

 

[e(ho0n3]

Like you said arildno, the cube will start slipping when F = f. Let u be the coefficient of static friction. f = umg so u = F/(mg). Why would I bother with the torque?

 

[arildno]

You need the inequality to ensure that the cube has NOT tipped prior to the point when it does begin to slide..