# Force applied to an inclined plane

Below is a picture of the problem. Any guidance would be helpful.

This problem isn't actually from any assignment, per se. I'm hoping that, by understanding this, it'll help me to understand a more complicated system where one inclined plane is resting on another inclined plane, and there's a mass on the top of the second plane, and the bottom incline is being accelerated by an applied force--but since I don't understand how this much simpler system works, I don't seem to have a chance at the more complicated system.

Forgot to include what the question is asking for. Since this isn't really a question I've been asked, I guess what I want to know is the acceleration of the system.

• Hint: What forces act on the block? Aug 13, 2014 at 18:04
• What question is the problem statement asking?
– CAF
Aug 13, 2014 at 18:35
• @CAF Ah, right, forgot to include that. It's asking for the acceleration of the system in terms of g, m, and theta. Aug 13, 2014 at 18:37

A shorter way to do this would be to use the concept of a pseudo-force.

When you observe the motion from the inclined plane's frame, you should be applying a pseudo-force (virtual force in the direction opposite to acceleration) because it is an accelerating frame.

Considering this, there are three forces on the block in this frame (which is seen stationery). One is gravity, downwards. Another is the normal reaction from the plane. Third is the pseudo-force, leftwards.

All you gotta do now is balance these three forces, because they cancel out and keep the block stationery.

• What is the force compensating $mg\sin \theta$ keeping the block stationary? The only possibility I see is a component of the applied force $F_A$ so that in order for the block to remain at rest, $F_A \cos \theta = mg \sin \theta$. Is that right?
– CAF
Aug 13, 2014 at 21:47
• In other words, I believe the component of the pseudo-force keeping the block stationary is $F_A \cos \theta$. If we go back to an inertial frame, the net force on the wedge is $F_A - mg \sin \theta \cos \theta = ma$ which agrees with one of the equations Addem found. But solving these two equations does not give $a=g\tan \theta$?
– CAF
Aug 14, 2014 at 8:12
• yes you are right Aug 16, 2014 at 19:11
• Can you please elaborate? I don't think I am correct given more thought and discussion elsewhere on the problem. I believe the equation $F_A \cos \theta = mg \sin \theta$ is wrong because there is no force keeping the block from sliding. It accelerates down the slope relative to the ground, but since the wedge moves, it remains in place. $F_A$ acts on the wedge, not block. The contact force between the two bodies is perpendicular to the wedge.
– CAF
Aug 17, 2014 at 8:39

Alright, I think I solved the problem: