Question

The mass of the wedge (movable wedge) is 7.89 kg the coefficient of friction is U b/w the 0.5 kg and the wedge for which the wedge just begins to move to the right when released from rest

Answer 1

@demitris

Answer 2

What's the question here?

Answer 3

we have to find U?

Answer 4

http://en.wikipedia.org/wiki/Friction#Dry_friction
Force down=\[0.4g\]=tension upwards
Normal force=\[0.5g \cdot \cos(37)\]
Tension upwards=normal force * U
\[0.5g \cdot \cos(37) U=0.4g\]

Answer 5

I'm oversimplifying, aren't I?

Answer 6

4 - T = 0.4a

T - mg sin 37 - U mg cos 37 = ma

Answer 7

Shuld it be like this ?

Answer 8

The only net force on the system is down, and since I"m assuming the block is resting on a (frictionless) table or something so it can't move in the y direction, that means that the forces on the system in the horizontal direction must cancel. If there's no net force, that means the momentum is constant -- it follows that the wedge will begin to move as soon as the masses do. Therefore, you just need to find the coefficient of friction such that the masses just begin to slide.

Answer 9

u mean .... mg sin 37 = U mg cos 37

Answer 10

Nope. Just equate the forces tugging at each end of the string (including friction)

Answer 11

0.5g sin 37= T- U*0.5*g cos 37...........(i)
T= 0.4*g