What should be the free body diagram?

[AlchemistK]

Homework Statement

The Question gives some data, and a figure to find out the acceleration.
In the arrangement shown (refer attachment) shown, M1=20kg, M2=30Kg, floor is smooth and friction between two blocks: μ(s) = 0.6, μ(k) = 0.4.
F= 400N
What are the forces acting on the two blocks?

Homework Equations

The Attempt at a Solution

On M1, there is F acting towards right, and the friction towards the left.
For M2, There is the friction acting towards right, and then the forces exerted by the pulleys.
In what direction and magnitude do they act?

 

[SammyS]

I'm assuming the pulleys are frictionless and have negligible mass.

Then the tension in the rope in all locations is F .

Forces exerted on the upper pulley: The rope exerts a force (equal to F) downward and a force, F to the left.

Forces exerted on the lower pulley: What do you think?

The pulleys may be considered as part of Block 1 .

https://www.physicsforums.com/attachment.php?attachmentid=40217&d=1319254345

 

[AlchemistK]

The force by the pulley on the lower block will be F towards the right, and F downwards.
Meaning that in the direction parallel to the floor, only the reaction of the friction force between M1 and M2 will act on M2,to the right, right?

 

[SammyS]

The force by the pulley on the lower block will be F towards the right, and F [STRIKE]downwards[/STRIKE]. upward
Meaning that in the direction parallel to the floor, only the reaction of the friction force between M1 and M2 will act on M2,to the right, right?

So the net force that the rope exerts on block #2, via the pulleys, is zero.

 

[asteriatic]

The free body diagram for this scenario would include the following forces:

1. Weight of M1: This would act downwards with a magnitude of 20kg x 9.8m/s^2 = 196N.

2. Normal force from the floor on M1: This would act upwards with a magnitude equal to the weight of M1, 196N.

3. Tension force in the string attached to M1: This would act towards the right with a magnitude equal to F, 400N.

4. Friction force on M1: This would act towards the left with a magnitude of μ(k) x normal force = 0.4 x 196N = 78.4N.

5. Weight of M2: This would act downwards with a magnitude of 30kg x 9.8m/s^2 = 294N.

6. Normal force from the floor on M2: This would act upwards with a magnitude equal to the weight of M2, 294N.

7. Tension force in the string attached to M2: This would act towards the left with a magnitude equal to F, 400N.

8. Friction force on M2: This would act towards the right with a magnitude of μ(s) x normal force = 0.6 x 294N = 176.4N.

Therefore, the free body diagram would show two blocks, each with four forces acting on them as described above. The direction and magnitude of each force would be labeled on the diagram.