Block on moving wedge (Maximum force for equilibrium problem

In summary: This leaves us with: F = μN([RF][/N]) cos(θ)Here, θ is the angle at which the block is released from the wedge, and [RF][/N] is the magnitude of the force applied by the block on the wedge. This gives us the equation for the maximum force that can be exerted on the block: Fmax = μN([RF][/N])cos(θ)This answer is correct.
  • #1
Omar Wali
14
0

Homework Statement



Given: The friction between the block with mass 11kg and the wedge with mass 22kg is 0.48. The surface between the wedge with mass 22kg and the horizontal plane is smooth (without friction). The [A][/g] = 9.8m/[/2]. A Block is released on the incline plane (top side of wedge).

What is the maximum force F which can be exerted on the 22kg block in order that the 11kg block does not move up the plane? Answer in units of N.

Homework Equations



ΣF = 0 when searching for an equilibrium point.

The Attempt at a Solution



I setup a FBD of the 11kg block. I listed the force of static friction, the normal force, the reactionary force F opposite the direction of the applied force F on the wedge and the force of gravity.

The calculations went as follows:

ΣF = 0

(μ*[F][/N]) + ([F][/RF]/cos(θ)) - ([F][/g]sin(θ) = 0

I solved for [F][/RF] and got ≈ 17.395 N. This answer is wrong and I'm not too sure where my error was. Thank you for the help in advance!
 
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  • #2
OK I'm obviously very new at writing things out not sure why things are striked out and the subscripts did not go in properly. I'll correct them next time. I believe it is still easy to read, however.
 
  • #3
Omar Wali said:

Homework Statement



Given: The friction between the block with mass 11kg and the wedge with mass 22kg is 0.48. The surface between the wedge with mass 22kg and the horizontal plane is smooth (without friction). The [A][/g] = 9.8m/[/2]. A Block is released on the incline plane (top side of wedge).

What is the maximum force F which can be exerted on the 22kg block in order that the 11kg block does not move up the plane? Answer in units of N.

2. Homework Equations

ΣF = 0 when searching for an equilibrium point.

3. The Attempt at a Solution

I setup a FBD of the 11kg block. I listed the force of static friction, the normal force, the reactionary force F opposite the direction of the applied force F on the wedge and the force of gravity.

The calculations went as follows:

ΣF = 0

(μ*[F][/N]) + ([F][/RF]/cos(θ)) - ([F][/g]sin(θ) = 0

I solved for [F][/RF] and got ≈ 17.395 N. This answer is wrong and I'm not too sure where my error was. Thank you for the help in advance!

The force F acts on the wedge, and the reactionary force is exerted by the wedge on the object that applies F. The sum of the forces applied on the block will accelerate it together with the wedge.

If both the wedge and the block accelerate together, the acceleration, multiplied by the total mass of the system, is equal to the external force F. The acceleration is horizontal, so the vertical components of the forces must cancel.
 

Related to Block on moving wedge (Maximum force for equilibrium problem

1. What is a block on moving wedge and what is its significance?

A block on moving wedge is a classic physics problem that involves a mass sitting on top of a wedge, which is then placed on a horizontal surface. The significance of this problem lies in its application to real-world situations, such as understanding the forces acting on an object on an inclined plane.

2. How is the maximum force for equilibrium calculated in a block on moving wedge problem?

The maximum force for equilibrium in a block on moving wedge problem can be calculated using the principles of statics. This involves analyzing the forces acting on the block and wedge, and setting them equal to zero in order to find the maximum force needed to keep the system in equilibrium.

3. What factors can affect the maximum force for equilibrium in a block on moving wedge problem?

The maximum force for equilibrium in a block on moving wedge problem can be affected by several factors, including the mass of the block and wedge, the angle of the incline, and the coefficient of friction between the surfaces.

4. How is the angle of the incline related to the maximum force for equilibrium in a block on moving wedge problem?

The angle of the incline is directly related to the maximum force for equilibrium in a block on moving wedge problem. As the angle of the incline increases, the maximum force for equilibrium also increases, since a steeper incline will require more force to keep the block and wedge in equilibrium.

5. Can the maximum force for equilibrium in a block on moving wedge problem ever be less than the weight of the block?

No, the maximum force for equilibrium will always be equal to or greater than the weight of the block in a block on moving wedge problem. This is because the weight of the block is one of the forces acting on the system, and in order to keep the system in equilibrium, the other forces must balance out the weight.

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