How Is the Force Calculated for Two Masses on a 30 Degree Incline?

[lpcampbe]

Two masses are being pulled up a 30 degree incline by a force F parallel to the incline. The velocity is constant and up the incline. The force is applied to a 200 kg mass and a string connects the 200 kg mass to a 150 kg mass. The coefficient of kinetic friction is 0.2. The force F = ?

So I drew a free body diagram, and it has F parallel to the inclined plane (Fx) and the normal force = Fy. There is mg, which is pointed directly down. The friction is pointed in the -Fx direction. I know F=uF_n, I also know that Fx = mgcos(theta) and Fy = mgsin(theta). Do I include the 2nd mass in my calculations? I'm assuming I do, in which case the mass is 350 kg. If the velocity is constant, acceleration = 0. Do I need to find the velocity to solve this problem?

 

[saltine]

I don't think you need to know the velocity. It is just there to justify using the kinetic friction coefficient for both masses.

 

[lpcampbe]

That's what I thought, but I have no idea how to do this problem. I used the Fx equation I listed above but it doesn't include the friction coefficient.

 

[glueball8]

draw a *FBD*

[tex] F_{y}=mg\times cos \phi [/tex]

[tex] F_{x}=F_{A}-mg\times sin \phi + F_{f} [/tex]

and [tex] F_{f}=\mu \times F_{N} = \mu \times mg\times cos \phi [/tex]

therefore

[tex] F_{A}=mgsin\phi + \mu \times mg\times cos \phi [/tex]

 

[lpcampbe]

Thanks, Bright Wang. That helped a lot. The Fa you used in that equation, is that the parallel force?

 

[glueball8]

applied force