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# Friction in Mechanics, how do I do this? watch

1. This is one question I haven't exactly done before:
A builder holds up a vertical piece of particle having weight 40 N by pressing the two sides, force P Newtons with his hands. If mu = 1/4 (0.25) what is the least value of P needed if the particle is not to slip.
I know, Friction = 1/2 W
So Friction = 20N
What do I do from here?
2. F=muR
3. That one liner is the one of the best
4. F = uR
(u = mew!)

= 0.25 x 40 = 10N

So in one direction (downwards) you have 40N.
Friction acts in the opposite direction, with a magnitude of 10N.
You want an overall force of magnitude 0N.
P will act in the same direction as the friction; it's stopping this 'vertical particle' (wtf XD) from slipping down.
So what is P?

Edit: I think I may be wrong.

First of all, the frictional force, is it per hand or what?
Secondly, I'm not sure that R is 40N. I think R = P, or something.

40 = P + 0.25R
R = P
40 = 1.25P
P = 32N

And if the frictional force is per hand, then:

40 = P + 9.5R
R = P
40 = 1.5P
P = 26.67N (2dp)

At least one of these has to be right.

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