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**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.

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

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.

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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