Two masses on inclined plane with frictionless pulley

[APPhysic]

Homework Statement

A 1.00-kg aluminum block and a 4.00-kg copper block are connected by a light string over a frictionless pulley. The two blocks
are allowed to move on a fixed steel block wedge (of angle 31.0°). The aluminum block is on the left side, on a flat plane. On the right is the copper block, sitting at a 31 degree angle. Picture is attached for clarification.
Coefficient of static friction for aluminum on steel: .61 Copper on steel: .53
Coefficient of kinetic friction for aluminum on steel: .47 Copper on steel: .36

Homework Equations

F=ma
ƩFx=Fg_x-tension-frictional force=(m₂)(acceleration)
Normal force=(massofcopperblock)(gravity)(cosθ)

The Attempt at a Solution

I was able to get the components of the force of gravity on the copper block: ≈33.6 N perpendicular to the plane, and ≈20.19 N parallel to the plane. Also, the normal force is 33.6 N. On the aluminum block, the normal and gravitational forces are both 9.8 N. From here, I am unsure how to incorporate the angle, the frictional forces, and how these will factor into solving for tension and acceleration of the bl

I seriously appreciate any help. It really means a lot for you to take the time to help a stranger out, I thank you immensely.

 

[ehild]

Hi APPhysic, welcome to PF

I see the equation for the cooper block, but miss the one for the aluminium block. What is its acceleration with respect to the acceleration of the cooper block? What forces act on it?

ehild

 

[APPhysic]

ehild,

Thanks for the welcome! I believe the aluminum block has tension, normal & gravitational force, and the force of friction acting on it. And won't the copper and aluminum block have the same acceleration or no?

 

[Panphobia]

Yes, the accelerations would be the same in this system.

 

[ehild]

ehild,

Thanks for the welcome! I believe the aluminum block has tension, normal & gravitational force, and the force of friction acting on it. And won't the copper and aluminum block have the same acceleration or no?

Yes, the acceleration of both blocks is the same parallel to the surfaces of the wedge. The normal force comes in by determining the force of friction. So what is the equation ƩF=ma for the aluminium block? .

ehild

 

[APPhysic]

Well, for the x component, ƩF_x=T-f_k=m₁a₁

And for y, ƩF_y=n-m₁g

So then, can I say that T-μ_km₁g=m₁a₁ ?

m₁ in this case being the aluminum block

 

[ehild]

Well, for the x component, ƩF_x=T-f_k=m₁a₁

And for y, ƩF_y=n-m₁g

So then, can I say that T-μ_km₁g=m₁a₁ ?

m₁ in this case being the aluminum block

Calculate the friction forces, write up the equations for both blocks. Add them, the tension will cancel, and you can get the acceleration.


ehild