# Given: coefficient kinetic and coefficient static frictions

1. Dec 26, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
A problem in engineering mechanics:
The blocks shown in the figure are connected by flexible inextensible cords passing over frictionless pulley. At Block A the coefficients of friction are Ms = 0.30 and Mk = 0.20 while at Block B they are Ms = 0.40 and Mk = 0.30. Compute the magnitude and direction of the friction force acting on each block.

the answer is Friction at A = 48lb
Friction at B = 36lb

2. Relevant equations
R is along the angle of friction w/c has two components:
Normal force at the vertical
Friction at the horizontal

3. The attempt at a solution
I do not know how to approach the problem since there are two given coefficient of friction and it seems very confusing.....

however i tried solving at block A using the coefficient of kinetic friction at A = 0.20:

F = friction
R = the force along the angle of friction

the angle of friction here lies along with R which is the coefficient of friction
tan(theta) = 0.20
theta = 11.3 degrees

so i used sine law:
$$\frac{\sin{79}}{300} = \frac{\sin{53}}{R}$$
R = 244lb

R has two components: the normal force N and the friction F
F = (244lb)(sin(11.3))
F = 48lb <<< at A

can u give some hints its very confusing

Last edited: Dec 26, 2007
2. Dec 26, 2007

### Staff: Mentor

I don't quite understand what you're doing. Where did the angle of 11.3 come from?

The first thing I'd do is figure out if this thing moves or not. Does it?

3. Dec 26, 2007

### Edwardo_Elric

no?

so ill try to use static friction