Submitted by jetay1 on Thu, 05/04/2017 - 21:00
I've been trying to get my head around the problem of the particle sliding down a moving wedge. I managed to get the required expression by using a pseudo force of $mA$ acting on the particle in the opposite direction to the motion of the wedge (where $A$ is the acceleration of the wedge), and then doing a horizontal force balance on both particle and wedge. Without the pseudo force I get \[A=\frac{a\cos\theta}{k}\] I know the pseudo force is something to do with the relative motion of the objects, but since it's not mentioned in the discussion below the question, I wonder if there's a better way to do the problem but with still using force balances?
I am ok with the rest of (i), but for question (ii), do we go back to the general case, or is $k$ still 3 and the particle still moving down a straight 45° line? The particle will obviously stay at rest when the inclined plane is flat enough, but it will be tricker to prove this in the general case, I think.
Actually, adding that extra
Actually, adding that extra term is very important. Newton's laws only hold in inertial frames (a frame of reference is inertial if it is not accelerating itself, or, equivalently, if a particle left alone in it moves with constant velocity). Since the wedge is itself accelerating, you need to work in the frame in which the smooth horizontal surface is at rest. Let's call this frame $S$. The acceleration $a^{\prime}$ of $P$ in $S$ is given by $$a^{\prime}=a\cos \theta -A,$$ where $A$ is the acceleration of the wedge in $S$ (in the direction of increasing height of the wedge). Hopefully I haven't confused you too much with that explanation. Frames of reference are very important when you study mechanics at university and start looking at special relativity.
For part (ii), in the case $\tan \theta \leq \mu$, $P$ is not in limiting equilibrium so I think it's safe to just write down what will happen.
Thanks
Thanks very much. There's nothing in any of the mechanics courses I've done that has involved relative motion, so I was a little unsure, but your explanation makes sense.