Coefficient of Friction in an inclined plane

In summary, the 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. The expression for the coefficient of friction µ in terms of the acceleration is μ = (g-a)/(g+a), where g is the acceleration due to gravity.
  • #1
e to the i pi
20
0
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




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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  • #2
e to the i pi said:
... 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
e to the i pi said:
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




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.

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
grzz said:
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
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
e to the i pi said:
...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
grzz said:
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
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.
 

1. What is the coefficient of friction in an inclined plane?

The coefficient of friction in an inclined plane is a measure of the resistance between two surfaces in contact when one surface is tilted or inclined. It is denoted by the symbol µ and has no unit of measurement.

2. How is the coefficient of friction in an inclined plane calculated?

The coefficient of friction in an inclined plane can be calculated by dividing the force of friction by the normal force. The force of friction is calculated by multiplying the coefficient of friction by the normal force.

3. What factors affect the coefficient of friction in an inclined plane?

The coefficient of friction in an inclined plane can be affected by factors such as the type of surfaces in contact, the smoothness of the surfaces, and the presence of any lubricants or adhesives.

4. Why is the coefficient of friction in an inclined plane important?

The coefficient of friction in an inclined plane is important because it helps determine the amount of force needed to move an object up or down the incline. It is also used in various engineering and design applications to ensure proper functioning of machines and structures.

5. How does the coefficient of friction in an inclined plane affect the motion of an object?

The coefficient of friction in an inclined plane affects the motion of an object by creating a resistance force that opposes the motion. This means that a higher coefficient of friction will require more force to move the object, while a lower coefficient of friction will result in easier motion.

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