Block on a moving inclined plane

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SUMMARY

The discussion centers on determining the horizontal force required to push an inclined plane so that a block of mass M remains stationary relative to it. The key equations involved include the gravitational force acting on the block, represented as m*g*sin(θ) for the component along the incline and m*g*cos(θ) for the normal force. The solution reveals that the horizontal force exerted by the ramp must equal the product of the mass of the ramp and g*tan(θ) to maintain equilibrium. This conclusion is reached through analyzing the forces acting on the block and the ramp's acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of trigonometry, particularly sine and cosine functions
  • Familiarity with free body diagrams and force resolution
  • Concept of normal force and its components in inclined planes
NEXT STEPS
  • Study the dynamics of inclined planes in physics
  • Learn about free body diagrams and how to resolve forces
  • Explore the concept of static equilibrium in mechanics
  • Investigate the effects of acceleration on forces in non-inertial reference frames
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to explain the concepts of forces on inclined planes and static equilibrium.

M4573R
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Homework Statement


A block of mass M is on a plane with an incline of Theta. What horizontal force must the ramp be pushed so that the block does not move relative to the ramp.

This is to be solved generally.

Homework Equations


Force of gravity on block: m*g*sin(theta)
Normal force: -m*g*cos(theta)

The Attempt at a Solution


I have no idea where I should start.
I the normal force needs to counteract the force of gravity on the block, but I don't know how the horizontal force on the ramp factors into the normal force.
 
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M4573R said:

Homework Statement


A block of mass M is on a plane with an incline of Theta. What horizontal force must the ramp be pushed so that the block does not move relative to the ramp.

This is to be solved generally.

Homework Equations


Force of gravity on block: m*g*sin(theta)
Normal force: -m*g*cos(theta)

The Attempt at a Solution


I have no idea where I should start.
I the normal force needs to counteract the force of gravity on the block, but I don't know how the horizontal force on the ramp factors into the normal force.

Welcome to PF.

You have 2 forces acting on the block.
Gravity acting vertically and the force of the ramp pushing horizontally against the block.
To keep the block from moving then the components acting along the plane of the incline must cancel.
The forces normal to the incline aren't useful for your purposes.
 
Is the force that the ramp is exerting on the block all in the normal force? And the normal force has a vertical and horizontal component?

After doing some trig I came to the answer that the ramp must push horizontally at m*g. Which happens to be the vertical force on the block. Is this correct? The normal Force keeps throwing me off.
 
M4573R said:
Is the force that the ramp is exerting on the block all in the normal force? And the normal force has a vertical and horizontal component?

After doing some trig I came to the answer that the ramp must push horizontally at m*g. Which happens to be the vertical force on the block. Is this correct? The normal Force keeps throwing me off.

No.

The m*g is acting along the incline with m*g*sinθ .

The force that the incline is exerting on the block is as the incline accelerates horizontally.
This translates into an m*a component up the ramp as a function of θ .
 
I'm just not sure what forces I have to cancel here. I've got a force down the slope, I've got a force of gravity pointing down, a normal force pointing perpendicular to the slope, and a force pushing horizontally on the block.
 
M4573R said:
I'm just not sure what forces I have to cancel here. I've got a force down the slope, I've got a force of gravity pointing down, a normal force pointing perpendicular to the slope, and a force pushing horizontally on the block.

Gravity acts downward.

It results in 2 components. Normal to the incline and parallel along the plane of the incline. That force is determined by the angle as Sinθ.

You're done with gravity.

Now the incline is accelerating at some rate if it is to exert a force. That force resolves itself into two components as well. The acceleration needed to keep the block steady creates an m*a*?θ that must equal the m*g*sinθ .
 
The accelerating ramp produces the horizontal force and the normal force of mg*cos(theta). Do I just need the normal force to equal the force down the slope?
 
M4573R said:
The accelerating ramp produces the horizontal force and the normal force of mg*cos(theta). Do I just need the normal force to equal the force down the slope?

No. I've already pointed out that the normal force component of gravity does not matter. m*g*cosθ is not useful.

"a" is the acceleration of the incline. It is the component of "a"*m that is parallel to the incline that you use.
 
I'm sorry that I'm not getting this yet. Soon or later its going to click. All my common sense tells me that the ramps horizontal force is acting along the normal of the block and that I need it to solve the problem.

If I have the force of the ramp as m*a:
Fblock = m*a, I can solve for a = Fblock / m.

You're saying this acceleration as the equal the acceleration of the block perpendicular to the ramp (g*sin(theta))?

As in
Fblock / m = g*sin(theta)
so the Force of the ramp = mg*sin(theta)?

Edit: Force on the ramp must = (mass of the ramp) * sin(theta)?

Edit2: Another try:
acceleration of the block = g*sin(theta)
horizontal acceleration of ramp = (Force of ramp / mass of ramp)
cos(theta) = ramps acceleration parallel to block's / horizontal acceleration of ramp
since acceleration parallel to block's must = g*sin(theta),
cos(theta) = g*sin(theta) / accel of ramp
Force of ramp = (mass of ramp * g * sin(theta)) / cos(theta)

Force on ramp = mass of ramp * g * tan(theta)

I don't know what else to try.
 
Last edited:
  • #10
M4573R said:
I'm sorry that I'm not getting this yet. Soon or later its going to click. All my common sense tells me that the ramps horizontal force is acting along the normal of the block and that I need it to solve the problem.

If I have the force of the ramp as m*a:
Fblock = m*a, I can solve for a = Fblock / m.

You're saying this acceleration as the equal the acceleration of the block perpendicular to the ramp (g*sin(theta))?

As in
Fblock / m = g*sin(theta)
so the Force of the ramp = mg*sin(theta)?

Edit: Force on the ramp must = (mass of the ramp) * sin(theta)?

Edit2: Another try:
acceleration of the block = g*sin(theta)
horizontal acceleration of ramp = (Force of ramp / mass of ramp)
cos(theta) = ramps acceleration parallel to block's / horizontal acceleration of ramp
since acceleration parallel to block's must = g*sin(theta),
cos(theta) = g*sin(theta) / accel of ramp
Force of ramp = (mass of ramp * g * sin(theta)) / cos(theta)

Force on ramp = mass of ramp * g * tan(theta)

I don't know what else to try.

That's just about it. I think the question wants it in terms of M. (The mass of the ramp/inclined plane is not given.)

It's probably easier to think in terms of the ramp accelerating, to determine the angle, because the actual force on the block balances when it is being accelerated in this fashion. That then makes the overall force on the ramp = to the acceleration of the ramp (is it massless, the problem doesn't say) and the acceleration imparted to the block of mass M.
 

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