# Friction and Force

1. Nov 9, 2009

### waldvocm

A block M=7.50kg is initially moving up an incline of 25.0degrees and is increasing sppeed with a= 4.44. The applied Force, F, is horizontal. The coefficients of friction between the block and incline are ms=0.443 and mk=0.312.

a) Fcos(25.0)-73.5sin(25.0)=0 F=34.3 This makes sense to me but does friction play a part in the applied force?

b) N=mgcos(25.0)+Fsin(25.0) N=81.1

c) fs=0.553*N

2. Nov 9, 2009

### Tzim

I don't quite understand what you are doing.In the direction of the incline the net force is not zero it is F-T=ma where T=(coefficient)*N
N=mgcos25

3. Nov 9, 2009

### Tzim

Sorry i have a mistake N=mgcos25+Fsin25

4. Nov 9, 2009

### Tzim

The force has has nothing to do with friction whereas the frixtion depends on the force because N the normal contact force depends on F in this excercise

5. Nov 9, 2009

### waldvocm

Thank you for the help

Now, I have the equations
F-T=ma
N=mgcos(25.0)+Fsin(25.0)
T=Coefficient*N

How do I solve them with all of the unknowns?

6. Nov 9, 2009

### Tzim

If you replace T with coefficient*N in the fisrt equation you have two equations with two unknowns.1)F-cof*N=ma
2)N=mgcos25+Fsin25