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Assume a small square block $m$ is sitting on a larger wedge-shaped block of mass $M$ at an upper angle $\theta$ such that the little block will slide on the big block if both are started from rest and no other forces are present. The large block is sitting on a frictionless table. The coefficient of static friction between the large and small blocks is µs. With what range of force $F$ can you push on the large block to the right such that the small block will remain motionless with respect to the large block and neither slide up nor down?

This question really is not too complicated without friction acting upon the small block (Preventing a block from sliding on a frictionless inclined plane)

But what happens when we add friction to the system? Why is it considered static, if at rest, the mass would slide down the plane?

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I wont solve the question for you, but I guess a clue will help you better to grab the concepts rather than some complicated solutions.

Here you see we will have to apply the external force F iff friction isn't enough to keep it from moving. So let the friction do the most it can, ie apply the maximum value of friction (coeff. Of friction times normal reaction) and the rest is done by an external force F.

The force F results in motion of the whole system and the acceleration would be force/(total mass) which results to some pseudo force on the smaller body, if you are sitting on the larger block (to be complicated in the same frame as of the larger block) some component of this force adds to the frictional force and keeps the smaller block from motion.

Figure out the rest yourself.

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Oh my god, this is an eleven month old question. I didn't noticed. – Abhishek Verma Jan 1 '14 at 18:03

Just add $\mu_k$ into the equations. See this for more information: enter image description here

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$m$=mass of block $M$=mass of plane

Both block and plane have the same acceleration

$A$=acceleration of plane $$F=(M+m)A$$

About to move down $$ mg \sin(x) - uN = mA\cos (x) $$ where $u$=coef. of static friction

$$ N - mg \cos (x)=mA \sin (x)$$

Solve for A and substitute in $F= (m+M)A$

About to move up $$ uN+mg\sin(x)=mA\cos(x)$$

Other equations remain the same. Solve for $A$ and then $F$ The two values of $F$ define the range of values

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