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Friction questions

A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is ms. The coefficient of kinetic friction is uk, with uk < us.

k and s are supposed to be subscripts.

Now I need to do:

Express all answers as functions of m, g, uk, us .

(A) Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force F must you be pushing the block just before the block begins to move?

(B) Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force?

(C) Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration a of the block after it begins to move.

Some clues would be great as I have gotten them wrong so far :frown: Thanks!
Reply 1
(A) While you are pushing the block, it isn't moving. So static friction is what governs the force(s) acting.

As you push against the block, and it isn't moving, then by the law of statics, the net sum of forces is zero. That means - the friction force must be the same as the force from your push. As you increase your push, the friction force, which will still be the same as your push, will increase, up to a limiting value. This limiting value is the value you get when you use the coefft of static friction, &#956;s.

The friction eqn is F = &#956;s.NR

Since the block is lying horizontally on the table, then NR = mg.
When the block is just about to move, then limiting friction has just been attained, you can use the friction eqn, and so ...

Ans: F = us.mg

(B) If you read through my answer to part (A), you will find it also helps you to answer (B).

(C) Let Fs be the force required to overcome static friction.
The block now moves, with the force Fs still being applied.
While moving, the force of friction is calculated using the coefft of kinetic friction. i.e. Fk = &#956;k.NR
Since the applied force is Fs and the frictional force is Fk, then what must be the net force?
Now use Newton's 2nd law to get the accln.

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