I want to demonstrate what force $F$ you would have to exert on an inclined plane of angle $t$, mass $M$ to prevent a block on top of it with mass $m$ from sliding up or down the ramp. I worked out an answer, but I figure it must be wrong because it doesn't factor the mass $M$ of the inclined plane into the force needed for the block on top to stay still.

Here is my logic:

  • The components normal to the force of gravity on the block on top are $mg \cos(t)$, $mg \sin(t)$
  • In particular, the component down the ramp is $mg \sin(t)$ as can be demonstrated with a visual
  • Thus, we want $F$ to have a component in the direction opposite the vector down ramp with equivalent force so that $F_{\rm ramp (net)} = 0$
  • So we want it to be true that $mg \sin(t) = F \cos(t)$
  • So $F = mg \tan(t)$.

Intuitively, this makes some sense: A steeper slope seems like it would require more force to counteract the component of gravity acting down the ramp.

However, this answer and explanation completely disregards $M$, the mass of the ramp itself.

Can somebody explain where I am going wrong, or if I obtained the right answer, why it is not dependent on $M$?


2 Answers 2


I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.

If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane so it accelerates at the same rate as the block, then the block won't move relative to the plane. What force on the plane is required to do this?

Assuming I've understood you correctly, the mistake you've made is to assume that the force you apply to the plane, $F$, is the same as the force the block exerts on the plane.


The force $f$ that the block exerts on the plane is not the same as the force $F$ that you exert on the plane.

When you apply a force F to the inclined plane it starts accelerating at some acceleration $a$ given by $a = F/(M + m)$. If you want the block to stay still relative to the plane the acceleration $a$ along the plane must be the same as the acceleration down the plane due to gravity:

$$ a \space \cos\theta = g \space \sin\theta $$


$$ \frac{F}{M + m} \space \cos\theta = g \space \sin\theta $$


$$ F = (M + m)g \space \tan\theta $$

  • $\begingroup$ Perfect explanation, thank you! If you dont mind me asking, why does the answer not depend on the mass of the little block $\endgroup$ Commented Sep 17, 2012 at 3:48
  • $\begingroup$ Because the force on the block is proportional to its mass i.e. $F_b = mg$ so the acceleration F/m is just $g$, the acceleration due to gravity. By contrast the force you apply on the plane isn't related to the mass of the plane so it's acceleration is $F_p/M$. The mass of the block cancels out but the mass of the plane does not. $\endgroup$ Commented Sep 17, 2012 at 5:49
  • $\begingroup$ I'll have to disagree. The correct answer is $(M+m)g\tan \theta$. The mistake you made is that you assumed that a force $F$ on the plane will produce an acceleration $F/M$ which is not true because the block is also applying an opposing force. $\endgroup$
    – Alraxite
    Commented Jan 22, 2013 at 10:24

The block and wedge have the same horizontal acceleration $a$ so use horizontal and vertical axes. FBD of $m$:

Y dir: $N \cos(t) - mg= 0$

X-dir: $N \sin(t)= ma$

Solving $a=g \tan (t)$

FBD of M: $F-N \sin(t)=Ma$

$$F-ma=Ma$$ $$F=(M+m)a$$ $$F=(M+m)g \tan(t)$$


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