00:01
All right, so not a free body diagram, but here is the situation that we're presented with.
00:06
You have a ramp incline at alpha degrees to the horizontal.
00:15
And then you have two masses, which are connected by a pulley, one on the ramp and one hanging from the string.
00:24
And the string is tension t, and tension is acting towards the polyolic number.
00:28
And recalling this mass on the incline of one mass and is hanging and two.
00:37
Okay, and so in part a, your velocity is acting uphill.
00:44
So your friction force will be acting opposite that, it will be acting downhill.
00:51
So you can resolve forces by making second law.
00:55
And we note, first of all, that here, from m2 t is equal to m2 g right the weight of the second mass is equal to the tension in the string okay and so then looking at the looking at the the y or oops the x direction you want no net force so this gives you the, let's see, the tension t, which is equal to m2g, is equal to the x component of weight 1.
01:52
So that's m1g sine alpha, that's the x component of m1g, plus f sub k, which is also acting downwards in this case.
02:02
And f sub k is just muu sub k times n right and n we can figure out from resolving forces in the y direction that will also be zero and n by the way is just this force here, the normal force.
02:38
And so n will be equal to m1g cosine alpha...