How to apply torque in this problem?

I am trying to solve the following question.

There are two wedges placed one on another on a horizontal plane as the picture given below. The coefficient of dynamic friction between each surface is $$\mu$$. The bottom wedge is given $$u$$ initial velocity. Find the final horizontal displacement of the upper wedge.($$\mathit{Note}$$:Measurements of the wedges are given)

I am trying to solve the question using Newton's second law $$F=ma$$. But the following situation puzzles my mind. After the bottom wedge has slipped a distance of half of its length, there will be a torque on the upper wedge caused by its weight and the normal reaction between the surfaces.

I am wondering how to apply this torque in the calculations. A guidance will be appreciated.

• The question is meaningless. The upper block will accelerate and its displacement will depend on a chosen time. Jul 5 at 14:58
• It is not clear to me what you are saying. So won't that torque affect the answer?
– ACB
Jul 5 at 15:53
• @R.W. Bird , please explain.
– ACB
Jul 5 at 23:32
• Displacement = Average Velocity x Time. What would you use for the time? Jul 6 at 13:23
• On second thought, see my answer below. Jul 6 at 14:30

You are correct in finding this exception. Whether this occurs will (obviously) depend on the actual values of $$M$$, $$m$$, $$\mu$$ and $$u$$, as well as the sizes of the blocks. By solving the initial differential equation, you can obtain a condition (inequality) for which this will never happen (where the relative velocity reaches zero before the slippage reaches half the length).

Therefore, if the above condition is not satisfied, the situation you described will occur. Actually, several situations could occur depending on the parameters and initial conditions. For example, we could have

1. The upper block loses contact completely before hitting the surface.
2. The rear end of the upper block hits the surface before the lower block comes to a stop. The lower block keeps traveling forward and eventually both blocks rest flat on the surface.
3. Same as above, but the lower block comes to a stop first, so that the upper block ends up in a sloped position.

I believe the best way to analyze the above is to work in the frame of the lower block. You have three dynamical degrees of freedom, all of which are functions of time:

1. The position and acceleration of the lower block
2. The angle of the upper block relative to the horizontal
3. The position and acceleration of the point of contact (pivot) along the length of the upper block.

In this frame, there will be four forces acting on the upper block:

1. Gravity
2. Friction from point of contact
3. Normal force from point of contact
4. Fictitious force

Do also take note that other things such as the moment of inertia will also be changing due to the changing pivot point. You also need a constraint equation for the blocks to remain in contact with each other. This is a non-trivial problem. You could first try to solve easier cases such as

1. Assuming the point of contact does not change after a certain time
2. Assuming the lower block is fixed in place, which enables you to focus entirely on the changing point of contact
• Thanks for your explanation. May be you are right, that first I could try to solve easier cases as you mentioned. But is there any possible way to solve the above question, ignoring that torque, such as 'energy conservation'?
– ACB
Jul 6 at 3:26
• @ACB Why will energy be conserved if there is friction? Jul 6 at 6:10
• Yes, some part of the energy will be expended to work against friction. But we can calculate it. I am referring to the work-energy relation.
– ACB
Jul 6 at 6:23
• @ACB in my opinion when the upper block will rotate about pivot (not necessarily at $l/2$) it (block) will gain some gravitational potential energy which again cannot be easily calculated. Hence, work energy theorem will not reduce calculations.
– Jay
Jul 6 at 8:46
• @ACB I suppose you could write down a Lagrangian with friction included for this case. Jul 6 at 21:58

Simplest solution: The lower mass m is struck by a horizontal impulse giving it a large momentum (mu). It leaves so quickly that the upper block does not gain a significant angular velocity. The upper block experiences a horizontal impulse from friction: μMgt = Mv where (t) is the contact time. It drops to the lower surface where it receives a similar (but negative) impulse from friction and stops. The displacement would be 2(v/2)t. And t = [L+(v/2)t]/u. You are right. Rotation would complicate the problem.