- #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.
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.
Last edited by a moderator: