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Coefficient of Friction in an inclined plane

  • #1
1. A box of mass 28 kilograms is held at rest on a rough plane inclined at an angle of 60° to the horizontal by a rope attached to a vertical pole. When the rope is cut the box moves down the plane with an acceleration of a m/s^2.
Find the expression for the coefficient of friction µ, in terms of the acceleration a m/s^2.




2. Here is a diagram of it:
http://img64.imageshack.us/img64/5774/dynamics.png [Broken]




3. I managed to split the 28kg force into a 28sin(60°) force going downwards and a 28cos(60°) force going leftwards.
Since 28sin(60°) = 28√(3)/2 = 14√(3) and 28cos(60°) = 28*1/2 = 14, there is a 14√3 kg force going perpendicular to the plane (downwards) and a 14kg force going parallel to the plane (leftwards).
I think friction stops the object from moving so the friction would be μ = 14kg, however I don't know how to find it in terms of the acceleration due to gravity.

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Answers and Replies

  • #2
993
13
... I managed to split the 28kg force into a 28sin(60°) force going downwards and a 28cos(60°) force going leftwards.


Since 28sin(60°) = 28√(3)/2 = 14√(3) and 28cos(60°) = 28*1/2 = 14, there is a 14√3 kg force going perpendicular to the plane (downwards) and a 14kg force going parallel to the plane (leftwards).
You are leaving out the acceleration due to gravity required to change the mass into weight.

Is not the 28x9.8sin60 downwards and parallel to the the plane?

Is not the 28x9.8cos60 downwards and perpendicular to the plane?
 
  • #3
PeterO
Homework Helper
2,425
46
1. A box of mass 28 kilograms is held at rest on a rough plane inclined at an angle of 60° to the horizontal by a rope attached to a vertical pole. When the rope is cut the box moves down the plane with an acceleration of a m/s^2.
Find the expression for the coefficient of friction µ, in terms of the acceleration a m/s^2.




2. Here is a diagram of it:
http://img64.imageshack.us/img64/5774/dynamics.png [Broken]




3. I managed to split the 28kg force into a 28sin(60°) force going downwards and a 28cos(60°) force going leftwards.
Since 28sin(60°) = 28√(3)/2 = 14√(3) and 28cos(60°) = 28*1/2 = 14, there is a 14√3 kg force going perpendicular to the plane (downwards) and a 14kg force going parallel to the plane (leftwards).
I think friction stops the object from moving so the friction would be μ = 14kg, however I don't know how to find it in terms of the acceleration due to gravity.

Homework Statement





Homework Equations





The Attempt at a Solution

What you think will happen is at odds with what they said does happen. You cannot make an assumption which contradicts the proposal.
 
Last edited by a moderator:
  • #4
You are leaving out the acceleration due to gravity required to change the mass into weight.

Is not the 28x9.8sin60 downwards and parallel to the the plane?

Is not the 28x9.8cos60 downwards and perpendicular to the plane?
So 14√3 * 49/10 = 343√3 / 5 Newtons going perpendicular to the plane (downwards).
And 14 * 49/10 = 343/5 Newtons going down the plane.
How do I work it out from here?
Would the formula F = ma help me?
If I let F = 343/5 and m = 28, then:
a = F/m so
a = (343/5) / 28
a = (343/140)
a = 49/20 m/s^2
But how do I calculate the friction?
 
  • #5
993
13
What does 49/(10) represent?
 
  • #6
Oh right... 9.8m/s^2 = 98/10 = 49/5. My mistake.
So 14√3 * 49/5 = 686√3 / 5 Newtons going perpendicular to the plane (downwards).
And 14 * 49/5 = 686/5 Newtons going down the plane.
How do I work it out from here?
Would the formula F = ma help me?
If I let F = 686/5 and m = 28, then:
a = F/m so
a = (686/5) / 28
a = (686/140)
a = 49/10 m/s^2 or in terms of gravity:
a = g/2
But how do I calculate the friction?
 
  • #7
993
13
...So 14√3 * 49/5 = 686√3 / 5 Newtons going perpendicular to the plane (downwards).
And 14 * 49/5 = 686/5 Newtons going down the plane ...
686√3 / 5 Newtons is going parallel to the plane (downwards).

686/5 Newtons is the normal reaction of the plane i.e. perpendicular to the plane.
Hence (coef of friction) x 686/5 is the frictional force parallel and up the plane.
 
  • #8
686√3 / 5 Newtons is going parallel to the plane (downwards).

686/5 Newtons is the normal reaction of the plane i.e. perpendicular to the plane.
Hence (coef of friction) x 686/5 is the frictional force parallel and up the plane.
I'm confused...
Where did you get that formula from?
And what is the 'frictional force'?
 
  • #9
Delphi51
Homework Helper
3,407
10
The formula is F down the ramp = mg*sin(60) and normal force Fn = mg*cos(60).
Since 28sin(60°) = 28√(3)/2 = 14√(3) and 28cos(60°) = 28*1/2 = 14
Just multiply each by 9.8 and you'll have 238 and 137 N for the forces down the ramp and into the ramp. If you don't want to go to numbers, keep the sine and cosine instead of the ugly radicals. Just my opinion!

The normal force Fn causes a friction force Ff = μ*Fn that opposes the motion down the ramp. Include it in your calculation of acceleration down the ramp. Of course you can't get a number for acceleration because you don't have a given value for μ. Your final answer will be solved for μ = some expression with unknown acceleration in it.
 

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