Applied Force on Superimposed Blocks

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Peppino
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1-kg Block 1 sits on 10-kg Block 2. Friction between the blocks is 0.4 static, 0.3 kinetic while the floor is frictionless. A 4-N force is exerted on Block 1. What is the acceleration of Block 2?

I know Block 1 won't move relative to Block 2 since it is canceled out by the frictional force, and I know Block 2 will be propelled by a frictional force. However, I am not sure which frictional forces come into play.

One way I see this is to treat both Blocks as one object, so that a 4-N force accelerates the 11-kg system at 4/11 = 0.364 m/s2. However, I'm not sure how to emulate this through each individual block in the free body diagrams.
 
Attachments
  • Diagram of two stacked blocks on a frictionless floor: block 1 of 1 kg on top, block 2 of 10 kg below, with static friction 0.4 and kinetic friction 0.3 labelled at their interface.
    block diagram.PNG
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  • Free-body diagram with labels: Ffs = 4 N pointing left, Fa = 4 N pointing right, Fn = 10 N pointing up, mg = 10 N pointing down.
    FBD top.PNG
    1.1 KB · Views: 464
  • Free-body diagram with a vertical upward arrow labelled Fn = 110N and a vertical downward arrow labelled (m1+m2)g = 110 N, a horizontal arrow to the right, and the note Ffs max = 44 N on the left with Ffs = 4 N written beside the horizontal arrow.
    FBD bottom.PNG
    1.6 KB · Views: 483
The blocks can move together if the force of friction is enough to accelerate the bigger one. When the blocks do not move with respect to each other, the static friction acts between them. Check if the maximum static friction is greater than the force needed to accelerate the big block.

ehild
 
But since there is no friction on the floor, wouldn't applying any force accelerate the big block?
 
Peppino said:
But since there is no friction on the floor, wouldn't applying any force accelerate the big block?

Yes, but not with the same acceleration as that of the small block.
You calculated the common acceleration of the blocks. What force is needed to accelerate the big block with a=0.364 m/s2?

ehild
 
That would be 3.64 N, so there is a 3.64 N frictional force? Where would that come from?
 
The blocks interact with the force of friction between them. It acts against the applied force in case of the 1 kg mass and it acts forward in case of the big block and accelerates it.

The coefficient of static friction is μs=0.4. You know that the actual force Fs can not exceed μs times the normal force: Fs ≤μsN. What is the Normal force between the blocks and what is the maximum of the static friction?


ehild