A system comprising blocks, a light frictionless pulley, and

[maestro3000]

Homework Statement

A system comprising blocks, a light frictionless pulley, and connecting ropes is shown. The 9-kg block is on a
smooth horizontal table (μ = 0). The surfaces of the 12-kg block are rough, with μ = 0.30.

In Fig. 5.7, the mass M is set so that it descends at constant velocity when released. The mass M is closest to: A) 2.4 kg B) 3.3 kg C) 3.0 kg D) 3.6 kg E) 2.7 kg

Homework Equations

F = MA
F_u = uN

The Attempt at a Solution

F = (m_1*g) - (m_2*g - u*m_3*g)
since the system is in equilibrium, F = 0
m_1 = m_2 - m_3*u
m_1 = 9 - (12)(.3)
m_1 = 5.4 kg

 

[maestro3000]

forgot to upload this

 

[RedDelicious]

Draw a free body diagram and then try to set up your equations again.

What forces are acting on the hanging mass, m1?

How about m2?

and then m3?

 

[maestro3000]

Draw a free body diagram and then try to set up your equations again.

What forces are acting on the hanging mass, m1?

How about m2?

and then m3?

the forces acting on m1 are T1 and m1g
the forces acting on m2 are T1, m2g, and the friction between the two blocks

 

[RedDelicious]

the forces acting on m1 are T1 and m1g
the forces acting on m2 are T1, m2g, and the friction between the two blocks

The first part is good. What equation do you get when you apply Newton's Second Law, F=ma, to that block using those forces?

There's a little bit of trouble here. In what direction do T1 and the friction force act? How about m2g? What do you get when you apply F = ma to this block using these forces?

 

[maestro3000]

The first part is good. What equation do you get when you apply Newton's Second Law, F=ma, to that block using those forces?

There's a little bit of trouble here. In what direction do T1 and the friction force act? How about m2g? What do you get when you apply F = ma to this block using these forces?

I got T1 = m1g
T1 and the friction force act in the same direction because of the frictionless pulley. I'm not sure if m2g affects the frictional force, because of the normal force being used

 

[RedDelicious]

I got T1 = m1g
T1 and the friction force act in the same direction because of the frictionless pulley. I'm not sure if m2g affects the frictional force, because of the normal force being used

Good on the first part.

Now think about this part carefully. As you initially said, the friction force is between the two blocks, and so the frictionless pulley doesn't factor into this. If two rough objects are being dragged across each other, will the friction force be in the direction of motion or opposite to it?

What are the forces in the y direction acting on m2? Is m2 accelerating in this direction? What does that tell you about them?