Friction problem from K&K

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Joined: Wed Oct 24, 2007 3:18 am

Friction problem from K&K

Postby niladri.bora » Wed Oct 24, 2007 3:20 am

#2.8 "A 4-kg block rests on top of a 5-kg block, which rests on a
frictionless table. The co-efficient of friction between the blocks is
such that the blocks start to slip when the horizontal force F applied
to the bottom block is 27 N. Suppose the horizontal force is now
applied only to the upper block. What is it's maximum value for the
blocks to slide without slipping relative to each other?"

Since the only contact surface with friction is that between the two
blocks and the normal force on this surface is the weight of the top
block, the force for slippage should be the same regardless of whether
it is applied to the top block or the bottom block. The answer given
in the book is 21.6 N. What am I missing?

Posts: 134
Joined: Thu Aug 30, 2007 10:21 am

Postby marten » Wed Oct 24, 2007 10:56 am

Let me take a stab at this, I think I see what is going on. First, the problem states that the maximum force applied to the bottom block before it starts slipping is 27N. Since the total mass of these two blocks is 9kg, the resulting acceleration for both blocks is 3 m/s^2. (no friction at the table) In order to accelerate the top block at this rate, the friction force between the two must be 4kg * 3 m/s^2 = 12N.

Now, if a force is applied to the top block, you are correct, exactly the same friction force exists between them. But now that same friction force must accelerate the bottom block, which is more massive. 12 N / 5 kg = 2.4 m/s^2. This is the maximum acceleration before the blocks start slipping. This total acceleration is achieved with a force of 2.4 m/s^2 * 9 kg = 21.6 N on the bottom block.

No need to actually calculate the normal force, or calculate mu.

Hope that makes sense.

Good luck!


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