Given: coefficient kinetic and coefficient static frictions

[Edwardo_Elric]

Homework Statement

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

Homework Equations

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

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

i am not so sure about this..
can u give some hints its very confusing

 

[Doc Al]

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?

 

[Edwardo_Elric]

no?

so ill try to use static friction