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# M1 Mechanics- question watch

1. Its easier to use an image then to recreate the image ....

Could anyone help me with this it would be greatly apreiated.

Thanks,
Bob
2. Resolve all the forces(including forces due to gravity) to make them perpendicular to one another, then, since you know that the resultant force is zero(due to the system being in equilibrium) and that Fmax = uR(since the particle is in limiting equilibruim), rearrange everything to find k.
3. ok thanks i'll try it.

I have already tried that way but i must of done something wrong :P
4. (Original post by ooboboo)
Its easier to use an image then to recreate the image ....

Could anyone help me with this it would be greatly apreiated.

Thanks,
Bob
Let's say A = alpha
F = kMgcosA + MgSinA
R = MgCosA + kMgSinA
F = mu(MgCosA + kMgSinA)
cancel off Mg
muCosA + k.mu.SinA = kCosA + SinA
k.mu.SinA - kCosA = SinA - muCosA
kCosA-k.mu.SinA = muCosA - SinA
k(CosA -muSinA) = muCosA - SinA
k = (muCosA - SinA)/(CosA - muSinA)

got there eventually.
5. thanks
6. thanks again i'v worked through it and don't quite understand why
F = kMgcosA + MgSinA becomes F = mu(MgCosA + kMgSinA) is this because of limited equilibruim?( resultant force = resistive forces)
7. (Original post by ooboboo)
thanks again i'v worked through it and don't quite understand why
F = kMgcosA + MgSinA becomes F = mu(MgCosA + kMgSinA) is this because of limited equilibruim?( resultant force = resistive forces)
Just to clarify (it's not that clear in the post), but when they said:
"F = kMgcosA + MgSinA
R = MgCosA + kMgSinA
F = mu(MgCosA + kMgSinA)"

The last F is Friction, and the first F is the force of weight and kmg acting down the plane.

As you probably noticed that though, the answer to your question is effectively yes - it's because it is on the point of slipping 'down the plane' (effectively limiting equilibrium, but it's important to note friction will be acting upwards as it opposes motion). Hence force down (first 'F') is equal to the maximum value of friction (second 'F' or μR).

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