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

Click For Summary

Homework Help Overview

The problem involves a block of mass (m) placed on a triangular wedge of mass (M) at an angle (α). The task is to determine the force (F) required to keep the smaller block stationary on the wedge, considering the coefficient of static friction (μs) and gravitational effects. The discussion includes the need for a free-body diagram (FBD) to analyze the forces acting on both blocks.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's laws, particularly the first and second, and the necessity of a free-body diagram to identify forces. There are attempts to resolve forces into components and establish equilibrium conditions. Some participants express uncertainty about the correct application of forces in the FBD.

Discussion Status

The discussion is ongoing, with participants sharing insights about the forces involved and the need for clarity in the free-body diagram. There is recognition of the need to resolve forces into components, but no consensus has been reached on the correct setup or solution approach.

Contextual Notes

Participants note the importance of documenting all relevant laws and theories, such as Newton's laws and the definition of friction. There is an emphasis on ensuring that all forces are accounted for in the FBD, which remains a point of confusion.

amajorflaw
Messages
2
Reaction score
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
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
 
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.
 
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.
 
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.
 
mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)

forces acting on m in equilibrium.
forces due to gravity = forces due to F.
 
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
 

Similar threads

  • · Replies 37 ·
2
Replies
37
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
20
Views
4K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 11 ·
Replies
11
Views
7K