# Inclined plane question [closed]

An object, mass $m$ is placed on an incline, angle $\theta$. System is at equilibrium. coefficients of static and kinetic frictions are $\mu_s$ and $\mu_k$ respectively. Then:

1) What is the Total contact force on the block by the surface??

2) Which are necessarily true?

a) $tan \theta$ is less than or equal to $\mu_s$

b) $f = \mu_s N$

c)$f= mgsin\theta$

My approach to 1)

total contact force will be $mgcos \theta$ (exerted by the incline as a normal reaction) + $\mu_s mgcos \theta$ (exerted by the incline in the form of friction) which are at right angles, result will be the square root of their squares, giving $mg cos \theta \sqrt{1+ \mu_s^2}$.

But the answer given is just mg!!! And the explanation is that total contact force = $mg cos \theta (= N) + mg sin \theta (= f)$ $\implies$ total force $= \sqrt{N^2+ f^2} = mg \sqrt{cos^2 \theta + sin^2 \theta }$

For 2)

I think all are right, except, $tan \theta = \mu_s$

but answer says b) is false!

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## closed as off-topic by Colin McFaul, Kyle Oman, DavePhD, Brandon Enright, Chris WhiteJun 12 '14 at 23:08

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Here we need to see that $$f\le N\mu_s= mg \cos\theta\ \mu_s$$

$N\mu_s$is the maximum friction that may be present in the surfaces,whereas here we need equilibrium.

The two forces along the incline must balance each other. So, $$f=ms\sin\theta \le \mu_s mg\cos\theta$$ If this inequality does not hold , ie. $$\tan\theta\le\mu_s$$

then the body will never remain in eqiulibrium. Now you can proceed !

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why less than equal to? why not equal to? –  Saurabh Raje May 23 '13 at 12:15
$mgsin \theta = \mu_s mg cos \theta$ –  Saurabh Raje May 23 '13 at 12:16
$\implies$ $tan \theta = \mu_s$ –  Saurabh Raje May 23 '13 at 12:16
Because it is the maximum friction that can be applied. See the friction curve , (that increase linearly and then becomes constant).$mg\sin \theta = \mu_s mg \cos \theta$ is not always true. $\mu$ is the surface property and independent of $\theta$ –  ABC May 23 '13 at 12:17
@SaurabhRaje hmmmm. for equilibrium $f=mgsin\theta$ agree? –  ABC May 23 '13 at 12:41