Block resting on slop with friction - statics

[y3ahright]

Homework Statement

a block of mass m resting on a 20° slope. The block has coefficients of friction µs = 0.80 and µk = 0.49 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass m2 = 2.0 kg.

What is the minimum mass m that will stick and not slip?

Homework Equations

Newtons second law sum of the forces = mass * acceleration

The Attempt at a Solution

I used the hanging block to find the tension to be equal to 9.8*(2) = 19.6 N

With that found I found that n = mgSin(20) and T = µs * mgSin(20)

So 19.6 = .8*m 9.8Sin(20) and m = 7.309 kg

The online homework says this is incorrect and I have no idea of any other way to go about this problem any help is appreciated.

Thanks

 

[Doc Al]

I used the hanging block to find the tension to be equal to 9.8*(2) = 19.6 N

Good.

With that found I found that n = mgSin(20) and T = µs * mgSin(20)

That's an incorrect expression for the normal force (wrong component of weight). And I assume you meant friction force, not T (for tension).

Hint: Three forces act on the block parallel to the incline.

 

[y3ahright]

Good.


That's an incorrect expression for the normal force (wrong component of weight).

I don't get what you mean by wrong component of weight, when i draw my FBD i have normal force going up and mg pointing down and

n - mg = 0 so n = mg but because its the 20 degree slope its n = mgSin20 ?

 

[Doc Al]

I don't get what you mean by wrong component of weight, when i draw my FBD i have normal force going up and mg pointing down and

While mg points down, the normal force doesn't point up--it's perpendicular to the surface, so it's at a angle.

n - mg = 0 so n = mg but because its the 20 degree slope its n = mgSin20 ?

It's not true that n = mg. To find the normal force, consider force components perpendicular to the surface--they must add to zero. What's the component of gravity perpendicular to the surface.