How Is Force Calculated to Maintain Position on an Inclined Wedge?

In summary: Sin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha) and solve for a. In summary, the force necessary to keep the smaller block at the same height on the wedge-shaped block is equal to the coefficient of friction multiplied by the gravitational acceleration.
  • #1
amajorflaw
2
0

Homework Statement


a block of mass (m) is placed on a bigger triangular shaped block with a mass (M). The wedge is of angle (α) . The coefficient of static friction between the two blocks is μs . What must the force (F) be in order to keep the smaller block at the same height on the wedge-shaped block? The answer is an equation in terms of the coefficient of friction, M, m, the angle and the gravitational acceleration. A FBD is also required. All laws must be documented and used properly.
clip_image002.jpg



Homework Equations


Newton's 1st Condition
Gravitational Acceleration
Definition of Friction


The Attempt at a Solution



I have the answer, but I'm having trouble getting to it.
The answer to the problem:
F = (M+m)((sinα-μscosα/cosα-μssinα))g
 
Last edited:
Physics news on Phys.org
  • #2
You are missing another important theory, Newton's 2nd. Also, you should draw a free-body diagram to clearly communicate all of the forces being exerted on this body.

Casey
 
  • #3
The acceleration of the system = a = F/(M + m )
the forces acting on the mass m are mg and ma. Resolve them into the components parallel and perpendicular to the surface of the wedge and find the condition for the equilibrium.
 
  • #4
Yeah that's where I'm stuck. I'm not sure where all of the forces apply on the two blocks, so my free body diagram is wrong. It's not the equations that are troubling me. It's getting all of the parts to the FBD that is troubling me.
 
  • #5
I'm not sure where all of the forces apply on the two blocks Both the blocks experience equal force in thr horizontal direction. You have to draw FBD for the body of mass m. As I have already mensioned, resolve them into the components parallel and perpendicular to the surface of the wedge and find the expression for a.
 
  • #6
mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)

forces acting on m in equilibrium.
forces due to gravity = forces due to F.
 
  • #7
mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)
That is correct. You can simplify this as g[Sin(alpha)-μsCos(alpha)]=a[Cos(alpha)-μsSin(alpha)]. Now substitute the value af a
 

What is gravitational acceleration?

Gravitational acceleration is the acceleration that an object experiences due to the force of gravity. It is the rate at which the velocity of an object changes when it is falling under the influence of gravity.

How is gravitational acceleration measured?

Gravitational acceleration is typically measured in meters per second squared (m/s²). It can be measured using devices such as accelerometers or by conducting experiments involving free-falling objects.

What is the value of gravitational acceleration on Earth?

The value of gravitational acceleration on Earth is approximately 9.8 m/s². This value can vary slightly depending on location and altitude, but it is generally considered to be a constant value for most everyday calculations.

How does gravitational acceleration affect objects of different masses?

Gravitational acceleration affects all objects equally, regardless of their mass. This means that all objects will experience the same acceleration due to gravity when falling towards the Earth. However, the force of gravity acting on an object will depend on its mass, with more massive objects experiencing a greater force.

How does gravitational acceleration vary on other celestial bodies?

Gravitational acceleration can vary on other celestial bodies depending on their mass and size. For example, the gravitational acceleration on the Moon is approximately 1.6 m/s², while on Jupiter it is approximately 24.8 m/s². This is due to the varying mass and size of these bodies, which affects the strength of their gravitational pull.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
369
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top