Acceleration of a block on a moving wedge

In summary, the conversation discusses a problem involving a block and a wedge on an inclined surface, where the wedge is acted upon by a horizontal force. The coefficient of static friction and the angle of the incline are given, and the goal is to find the maximum and minimum values of force for which the block does not slip. The conversation delves into the forces acting on the block and how to incorporate the acceleration of the wedge into the solution. Eventually, it is determined that while there is zero acceleration along the incline, the block and wedge are both accelerating in an inertial reference frame. This is known as d'Alembert's principle.
  • #1
fontseeker

Homework Statement



A block of mass 0.5Kg rests on the inclined surface of a wedge of mass 2kg. The wedge is acted on by a horizontal force that slides on a frictionless surface. If the coefficient of static friction between the wedge and the block is 0.8 and the angle is 35 degrees, find the maximum and minimum values of F for which the block does not slip.

media%2F1ef%2F1ef9fdd3-f706-44e4-b5cd-d048423698e0%2FphpYjmyer.png


Homework Equations



F = ma

The Attempt at a Solution



IMG_4450.jpg


I don't fully understand how to apply the acceleration of the wedge to the block. I know both objects are traveling with the same acceleration, but then wouldn't that mean that the block is moving with an acceleration down the ramp?
 

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  • #2
The way you have drawn your sketch implies that ay lifting off from the wedge and ax accelerating down the wedge is not what is happening. In fact, there is no acceleration between the block and wedge.
Try starting to look at the block and the wedge when there is no force pulling the wedge. Think about the following:
  • What forces are acting on the block?
  • What is the amount of frictional force necessary to keep it in place?
  • How much additional Normal force is required to get the necessary frictional force.
  • When the external force is applied, how does that transfer to additional forces applied to the block?
 
  • #3
Sum forces up and down the ramp and set =ma where m = mass and a = accel. of block. Do this twice, considering the friction force could act as to push the block either up or down the ramp

Hint: F could be negative!
.
 
  • #4
rude man said:
Sum forces up and down the ramp and set =ma where m = mass and a = accel. of block. Do this twice, considering the friction force could act as to push the block either up or down the ramp

Hint: F could be negative!
.
I've done the summation of forces uses the incline as my x-axis for both friction going up and down. However, how would I incorporate the acceleration of the wedge?
 
  • #5
fontseeker said:
I've done the summation of forces uses the incline as my x-axis for both friction going up and down. However, how would I incorporate the acceleration of the wedge?
Remeber ΣF = ma? What's a along the incline & how does it relate to F?
 
  • #6
rude man said:
Remeber ΣF = ma? What's a along the incline & how does it relate to F?
Wouldn't a along the incline have to be 0 in order for the block not to slide?
 
  • #7
fontseeker said:
Wouldn't a along the incline have to be 0 in order for the block not to slide?
I'm changing my answer somewhat.

Yes, with respect to the incline there is zero acceleration - zero motion in fact. But the equation ΣF = ma applies in an inertial, not accelerating, coordinate system. In an inertial refrence system the block and wedge are both accelerating = a. But you can sum forces in the direction of the incline and let those sum to m times acceleration of the block in the direction of the incline but realize that acceleration along the direction of the incline is not the same as acceleration along the directon of a.

(In fact you can think of F as being replaced by another force F' = -ma, then the block experiences zero net acceleration so now ΣF' = 0, F' = F - ma. This is known as d'Alembert's principle which you may run into in a later dynamics course.)

A similar scenario is say you're in an elevator & the cables broke & you're dropping with acceleration -g. If you were to try to drop an item from your hand the item would have no motion with respect to you but of course you & the item are both accelerating at -g in inertial space. Same with the block and the wedge.
 
Last edited:

1. How does a moving wedge affect the acceleration of a block on top of it?

The moving wedge will affect the acceleration of the block by providing an additional force that acts on the block in the direction of motion. This force is known as the normal force and is responsible for the acceleration of the block.

2. What factors affect the acceleration of the block on a moving wedge?

The acceleration of the block on a moving wedge is affected by the angle of the wedge, the mass of the block, and the coefficient of friction between the block and the wedge. These factors determine the magnitude of the normal force and therefore, the acceleration of the block.

3. Does the acceleration of the block change as the wedge moves?

Yes, the acceleration of the block will change as the wedge moves. This is because the angle of the wedge changes, affecting the magnitude of the normal force and therefore, the acceleration of the block.

4. How can I calculate the acceleration of a block on a moving wedge?

The acceleration of the block on a moving wedge can be calculated using Newton's second law of motion (F=ma). The net force acting on the block is the sum of the gravitational force and the normal force, which can be determined using trigonometric functions and the given parameters.

5. What is the role of friction in the acceleration of a block on a moving wedge?

Friction plays a crucial role in the acceleration of a block on a moving wedge. It determines the maximum possible acceleration of the block and can either assist or oppose the motion, depending on the direction of the force. The coefficient of friction also affects the magnitude of the normal force and therefore, the acceleration of the block.

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