Frictional Forces and Acceleration: Solving for Horizontal Component

AI Thread Summary
The discussion focuses on calculating the horizontal component of the frictional force between two blocks on a frictionless surface. The applied force of 1.2 N and the mass of the lower block (1.0 kg) lead to a shared acceleration for both blocks. Participants analyze the forces acting on each block using F = ma, concluding that the frictional force on the smaller block is 0.4 N. The calculations involve rearranging the equations to isolate the frictional force based on the total mass and acceleration. Clarification and guidance are requested to ensure the solution is correct.
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Homework Statement



Two blocks are accelerated across a horizontal frictionless surface as shown. Frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. F = 1.2 N and M = 1.0 kg

Homework Equations



what is the horizontal component (frictional force) of the force of the large block on the small block?


The Attempt at a Solution



 

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If the acceleration is the same, then the force must change given the masses are different.

Given F = ma

For the lower block, F1 = 2M*a

For the upper block, F2 = M*a

Where M = 1.0kg and F1 = 1.2N

Given a is equal, you can rearrange the two and solve for F2

Please note you are really supposed to supply an attempt at the solution.

Jared
 
I know the answer is 0.4 N to the right. I don't know how it gets. But here's what I'm thinking.

i think the given F applies to both blocks, so

F= ma
1.2 = 3M * a
a = 0.4

then F1 = ma = 1 * 0.4 = 0.4 N

I don't know if it makes sense. Can someone give idea and guidance on how to solve this problem?
 
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