# Troubling frictional force problems

1. Oct 7, 2006

### OVB

The first problem says that a 68 kg crate is dragged across a floor by pulling a rope at a 15 degree angle to the ground.

1. If the static friction coefficient is 0.5, what is the minimum tension needed to start the crate's motion?

I did Tcos15 = 0.5(mg) and got 344.95, but my book says 300 N

2. If kinetic frictional coefficient is 0.35, what is the magnitude of the initial crate's acceleration?

I did Tcos15 - 0.35mg = ma and got 1.5, but the book says 1.3 m/s/s

The second problem has a mass A weighing 44 N connected to a pulley (no friction or mass). The pulley connects to a mass B (weighing 22 N) in perependicular to the mass A (B is hanging off the side fo the table). There is a weight for C that will keep the combined A + C mass from sliding if the static friction coeff. is 0.2 (I got that part, it's 66 N). The next part asks:

If mass C is suddenly lifted off A, what is the acceleration of block A if the kinetic frictional coefficient between A and the table is 0.15?

I did 22 N - (.15)(44N) = Ma where M = 4.49 kg and got 3.4 m/s/s, but hte book says it is 2.3....

The last difficulty I have has to do with the difference between static and kinetic friction when a force is directed at an angle. If a mass has a downward force acted upon at a certain angle (in my case I am referring to a force vector in the fourth quadrant, with theta between the third quadrantal line and the force vector) I can easily find the kinetic friction by doing Fsin8 - (Uk)(Fcos8 + mg) = 0 where 8 represents theta. I know this because the perpendicular force component in this case is Fcos8, so the combined actual weight plus the applied force is equal to the normal force. However, for static frictional force, it seems that for some reason all is done (according to my book) that Fsin8 = (Us)(Fcos8). Why is the mg actual weight not accounted for in this calculation for the normal force?

I know this is a lot to ask for, but if I even got piecewise feedback that would be greatly appreciated.

2. Oct 7, 2006

### OlderDan

It appears you have neglected the upward force on the crate from the rope. This reduces the normal force and hence the frictional force. If you fix this problem here, you can probably do the other questions.

3. Oct 7, 2006

### OVB

but how does this do that? I know it's a rule, but in terms of vectors how does it make sense? I draw the upward force vector and the normal force vector together; I know the net force is the difference between the two, but why aren't the vectors added?

Thanks for the help though.

4. Oct 7, 2006

### PhanthomJay

They are. Tsin 15 + N =mg;
N=mg-Tsin15