What is the angle at which an object on an incline plane begins to slide?

Click For Summary

Homework Help Overview

The problem involves determining the angle at which an object on an inclined plane begins to slide, given its mass and the coefficient of static friction. The context includes forces acting on the object, particularly static friction and gravitational components.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between static friction and the normal force, questioning the correct application of formulas. There are attempts to clarify the components of forces acting on the object, particularly the normal force and its relation to the incline.

Discussion Status

The discussion has progressed with participants exploring the conditions for the object to begin sliding, including setting up equations involving sine and cosine functions. Some guidance has been offered regarding the relationship between forces, but no consensus has been reached on the final angle calculation.

Contextual Notes

There is mention of potential missing information and the need for clarity on the normal force in relation to the incline, indicating that assumptions about the setup may be under discussion.

pb23me
Messages
204
Reaction score
0
1. The problem statement, all variables and g
object on an incline plane has a mass of 10 kgs and the coefficient of static friction is 0.35 at what angle does the object begin to slide.

Homework Equations


Fnet=ma Ff=sin(theta)(mg)


The Attempt at a Solution

I don't come close to the solution because i feel as though some piece of information is missing. Therefore i can't begin to solve the problem.
 
Physics news on Phys.org
What's the maximum value of static friction for a given angle?
 
isnt it Fn(Us) or normal force times static friction?
 
not actuallu u that symbol that looks like a u the micro symbol
 
so the normal force is equal to (mg) or in this case (10kg)*(9.86) oh ok i think I've got it now thank you very much...
 
pb23me said:
isnt it Fn(Us) or normal force times static friction?
The maximum value for static friction equals μFn.

pb23me said:
so the normal force is equal to (mg) or in this case (10kg)*(9.86) oh ok i think I've got it now thank you very much...
Careful! The normal force does not equal mg. (It would if the surface were horizontal, but it's not.) Hint: Find the component of the weight perpendicular to the surface.
 
cos(theta)*mg
 
pb23me said:
cos(theta)*mg
Good!

So what's the maximum static friction force?
 
(.35)*cos(theta)*mg
 
  • #10
pb23me said:
(.35)*cos(theta)*mg
Good.

So what's the condition for the object to just barely start to slide?
 
  • #11
is this correct?
 
  • #12
sin(theta)mg be greater than (.35)*cos(theta)*mg
 
  • #13
pb23me said:
sin(theta)mg be greater than (.35)*cos(theta)*mg
Yes. But to solve for the point where the object is just about to slide, set them equal. That will be the critical angle.
 
  • #14
ok i got tan(theta)=.35 inverse tan(.35)= 19.3 deg
 
  • #15
pb23me said:
ok i got tan(theta)=.35 inverse tan(.35)= 19.3 deg
Sounds good to me.
 
  • #16
awsome thank you soooo much!
 

Similar threads

Replies
11
Views
1K
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
22
Views
2K
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K