Question

how to solve the questions related to friction of two blocks, when force acts on upper body and friction acts only on the upper side of lower body....

Answer 1

Block on a block problem, with frictionup vote9down votefavorite9

Consider two blocks, one on top of the other on a frictionless table, with masses m1m1 and m2m2respectively. There is appreciable friction between the blocks, with coefficients μsμs and μkμk for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force FF (say, to the right) that can be applied to the top block so that the two blocks accelerate together.

The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with xx oriented to the right), the sum of horizontal forces for the top block is

F−Fsf=m1aF−Fsf=m1a

and for the bottom block

Fsf=m2aFsf=m2a

where FsfFsf is the force of static friction. Solving for aa in these two expressions, and then equating them, gives

F=(m1+m2)Fsfm2F=(m1+m2)Fsfm2

The maximum such force will therefore be achieved when FsfFsf is maxed out at μsm1gμsm1g, so

Fmax=m1m2μs(m1+m2)gFmax=m1m2μs(m1+m2)g

I understand this solution, but conceptually I don't have a response to the following nagging question: FmaxFmax is clearly larger than the max static friction force μsm1gμsm1g (because m1+m2m2>1m1+m2m2>1), so why doesn't the application of a force of magnitude FmaxFmax to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force FF of magnitude greater than μsm1gμsm1g would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.

newtonian-mechanics forces frictionshareciteimprove this questionedited Jan 17 '13 at 7:47Qmechanic♦89.2k11146829asked Jan 17 '13 at 3:15symplectomorphic188117That would be true if the bottom block were held fixed. But in the end you're applying the force to both blocks (although it's really touching the top block), and the bottom block does accelerate. So when you look in that reference frame, you're still below the maximum static friction. – Chris Gerig Jan 17 '13 at 3:26Can you elaborate a bit, or spell this out exactly? I don't know what it means to say "when you look in that reference frame, you're still below the max static friction." I'm thinking about free-body diagrams. Consider the opposite case, where a force FF of magnitude less than the max static friction force is applied. Then the sum of horizontal forces on the top block gives zero acceleration, but the reaction friction force on the bottom block gives a positive acceleration -- & yet shouldn't the blocks be accelerating together in that case? I want a way to be totally rigorous about this. – sympl