# Accelerated wedge [closed]

Consider the following figure, where a mass $m$ is placed on a wedge which is accelerated to the right ($\vec a$) which could be negative. The coefficient of friction between the mass and the wedge is $\mu$.

I sketched the 3 forces I know that the mass "feels" but I don't know that happens when the wedge is accelerated. Could any one give me a hint for this?

Is this $m\vec a$ vector I sketched in red correct? ## closed as off-topic by John Rennie, Danu, ACuriousMind♦, Ali, JamalSSep 28 '14 at 15:13

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## 1 Answer

You're totally correct with the m*a vector! Since the wedge is accelerated and we're assuming that the mass is sort of stuck to it (moves on its surface but stays with the wedge), in a frame of reference attached to the wedge it's going to experience a fictitious inertial force (imagine sitting next to that mass; the bigger the acceleration, the more you'd be pushed into the block).

It seems to me that since the acceleration of the wedge is constant (is it?), then by doing the classic separation of forces between directions parallel and perpendicular to the wedge's surface you'd be able to get to the equations of motion of the mass, which I'm assuming is what you need. Some geometry is pretty much all you need right now (assuming you know the angle between the wedge's slanted surface and the ground level).

• Thanks a lot. I simply need to be sure my reasoning is correct. And yes, I have the inclination angle of the wedge. – E Be Sep 28 '14 at 9:35
• Awesome, glad to hear that :) – Perfi Sep 28 '14 at 9:41
• Making more progress on this exercise, I am trying to find the range for which the block remains at rest with respect to the plane (the wedge). So I came with the following: $$\vec a =\vec g \cdot \dfrac{\sin \theta -\mu \sin \theta }{\cos \theta + \mu \sin \theta}$$. Given that the plane is inclined in angle $\theta$. Now, where is the range, i.e how I write the answer in the form $$\vec a \in (?,?)$$ – E Be Sep 28 '14 at 10:00
• I'm not sure that derivation you have there is correct... could you expand upon your reasoning about it? That coefficient of friction you have is the maximum value. If you apply little pressure to the mass, the force of friction is going to be small, right? You should look into when do the forces of inertia and gravity on the mass along the parallel surface perfectly counteract the maximum friction force you have there. Then, for all smaller forces (this is going to depend on the wedge's acceleration) the force of friction is enough to prevent the mass from moving. – Perfi Sep 28 '14 at 10:42