How Do Newtonian Forces Affect the Interaction Between Two Blocks?

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SUMMARY

The discussion focuses on the interaction between two blocks, A (3kg) and B (1kg), under the influence of Newtonian forces. When block B is released, block A is propelled to the right with a force of 160N, while static friction between the blocks is 0.50. The normal force exerted by block A on block B is calculated to be 40N, and the maximum force of friction on block B is determined to be 20N. The vertical force exerted by block B on block A requires further clarification regarding the dynamics of friction and gravity.

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  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Familiarity with normal force calculations
  • Basic principles of force equilibrium
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  • Learn about the implications of static friction in multi-body systems
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Nicolaus
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Homework Statement


Block B (1kg) is held in contact with block A (3kg). When block B is let go, A is pushed to the right with a force of 160N. Static friction is .50 between the 2 blocks.
Block A pushes on B such that B is not touching the surface upon which A is moving on. It's held in the air with only the normal force from A supporting it.
Find the force of friction on Block B, the normal force exerted by A on B, and vertical force exerted by B on A.

Homework Equations


Newtonian forces of friction, gravity, etc

The Attempt at a Solution


For the Normal force of A on B:
to find a: Fnet = 160N = m(A+B)a --> a = 160/4 = 40m/s^2
mAa = F - F(BonA)
FBonA = 160 - (3x40) = 40N = FAonB from Third Law

For Force of Friction on B: Since friction would be working against gravity and there is no vertical motion, wouldn't the force of friction be mBg, its weight?
The Max Ff would be Ff=μN = .50(40)=20N.
Not sure about the vertical force of B on A.
 
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Nicolaus said:
For Force of Friction on B: Since friction would be working against gravity and there is no vertical motion, wouldn't the force of friction be mBg, its weight?
How can friction oppose gravity - isn;t this the normal force or did you leave out part of the description?
 

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