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

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pb23me
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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.
 
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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.
 
pb23me said:
cos(theta)*mg
Good!

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

So what's the condition for the object to just barely start to slide?
 
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.
 
pb23me said:
ok i got tan(theta)=.35 inverse tan(.35)= 19.3 deg
Sounds good to me.