# Follow up question on a block on an accelerating inclined plane

A follow-up on this post: How can an accelerating inclined plane prevent a block on it from sliding?

"If F is larger than in the no relative movement condition then the magnitude of N2 is larger and the block accelerates up the inclined plane."

If the block moves up (and to the left), it means there is a left horizontal net force. Where does this force come from? Because horizontal component of normal force always points to the right, I feel the block should always moves down.

If the block moves up (and to the left), it means there is a left horizontal net force.

An inertial observer would say the block is not moving to the left. From the ramp frame it is moving to the left, but from the lab frame it is accelerating to the right. It is simply not accelerating to the right as quickly as the ramp is.

When a body accelerates, a pseudo force acts on it. Pseudo force acts opposite to the accelerated body. However, we use pseudo force only when we take the frame of reference as the moving body itself.

For example, if you are in a fast moving vehicle, you feel like you are getting pulled in the backward direction, even though the only force that seems to be acting on you is gravitational force. This is why it goes upwards.

In the problem, as there is a force on the inclined plane towards the positive x axis, the pseudo force acts in the negative x axis.

(reference as ground): they wanted you to visualise it was like, if the individual acceleration of inclined plane is more than individual acceleration of the block, it would go in the leftwards direction, as the wedge is faster than the block.

(reference point as moving inclined plane): if pseudo force is Fp and if F is the horizontal force of inclined plane, and F2 is the net horizontal acceleration on inclined plane for the block, F2 = F - Fp = Ma - ma

This is what i visualise:

If incline of inclined plane is Θ, then we can write equation of motion along incline of inclined plane as:

FpcosΘ - FgsinΘ = mA

where Fp = ma

Fg = mg

A is acceleration of block with mass m along the slant of inclined plane

a is the acceleration of inclined plane

this makes the equation: m(acosΘ-gsinΘ) = mA

Therefore, you can see, that if acosΘ > gsinΘ, A will be positive if acosΘ < gsinΘ, A will be negative.

When A is positive, it moves up the plane, when A is negative it moves down the plane.

Please let me know if you have any queries related to my answer, as this is my first answer and i am not sure whether i have formatted my answer correctly.

• MS Paint is a very beloved tool, but the use of MathJax is also encouraged around these parts. Welcome! Commented Jan 25, 2023 at 17:28
• Thanks for the diagram and calculation. Commented Jan 25, 2023 at 23:09