A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 5.8 N and a vertical force P are then applied to the block. The coefficients of friction for the block and surface are mus = 0.45 and muk = 0.22. (a) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 8.0 N. N (b) Determine the magnitude of the frictional force acting on the block if the magnitude op P is 10.0 N. N (c) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 12.0 N. N

Explanation / Answer

First, determine the normal force Fn, which is the upward force that the surface exerts on the block.
To do this, notice that the upward and downward forces balance (otherwise the block would be accelerating vertically, which it's not).
Upward forces: Fn + P
Downward forces: mg (i.e., the weight)

They're in balance, so: Fn + P = mg
or:
(1): Fn = mg - P

Now hold that thought. Next, consider the sideways forces:
Pointing right: 5.8 N
Pointing left: Friction (F_f).

We don't know how much F_f is yet, because we don't know whether we should use s or k. So let's figure that out.
The _maximum_ amount of friction force that s can supply is: (Fn)(s) = (mg - P)(0.45)
= (2.5*9.8 - 8)(0.45) = 7.43 N

Since the rightward force is less than this (only 5.8N), the maximum static friction is more than enough to prevent the block from sliding.

In cases where the block is not sliding, the _actual_ friction exactly matches the opposing force, which is 5.8 N. So this means the magnitude of the frictional force is 5.8 N

b) The _maximum_ amount of friction force that s can supply is: (Fn)(s) = (mg - P)(0.45)
= (2.5*9.8 - 10)(0.45) = 6.53 N

Since the rightward force is less than this (only 5.8N), the maximum static friction is more than enough to prevent the block from sliding.

In cases where the block is not sliding, the _actual_ friction exactly matches the opposing force, which is 5.8 N. So this means the magnitude of the frictional force is 5.8 N

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