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Inclined plane

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1. A mass of 4 kg lies on a rough plane which is inclined at 30 to the horizontal. A light string has one end attached to this mass, passes up the line of greatest slope, over a smooth pulley fixed at the top oc the plane and carries a freely hanging mass of 1kg at its other end. The tension in the string is just sufficient to prevent the 4kg mass from sliding down the slope. Find the coefficient of friction between the 4kg mass and the plane.
2. Where are you up to?

If you're yet to begin, I'd advise first drawing a diagram.
3. (Original post by Kozmo)
Where are you up to?

If you're yet to begin, I'd advise first drawing a diagram.
I've tried to solve it through resolving and f=am but am still left with two variables to solve
4. (Original post by Gemma_98)
I've tried to solve it through resolving and f=am but am still left with two variables to solve
Is this the entire question? I can't help but feel as though a little information is missing despite coming to a solution.
5. (Original post by Kozmo)
Is this the entire question? I can't help but feel as though a little information is missing despite coming to a solution.
That's the entire question
6. (Original post by Gemma_98)
That's the entire question
Alright, well I approached it by thinking about what is acting on which:

The mass on the end of the wire is acting on the object of 4kg - this is providing a 'counter-weight' in essence due to the tension force, T. F=mg.

The mass acting downwards is 4kg, we can change this to a force using F=mg.

We then want to find friction force, F.

We must then work out the horizontal component of the weight (as this is the same direction as the friction force and tension force). - this is done by resolving the force.

Once we have done this it is a matter of balancing the equation:

Tension force, T + friction force, F = horizontal component of weight, mg

as we know it's in equilibrium and thus has a zero resultant force.

7. This is a diagram of what we have, try work it out from here. If you're having any problems, I'm happy to assist where needed

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Updated: April 19, 2016
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