What will the FBD of 2 blocks being pulled look like?

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SUMMARY

The forum discussion centers on the free body diagrams (FBD) of two blocks, A and B, where Block A rests on Block B, and Block B is being pulled at a constant velocity. Both blocks have equal mass and size, and friction exists between Block B and the table. The forces acting on Block A include its weight and the normal force from Block B, while Block B experiences gravitational force, the normal force from the table, an applied force, and kinetic friction opposing the motion. The discussion emphasizes the need to clarify whether there is friction between the two blocks and how to represent these forces accurately in FBDs.

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  • Familiarity with concepts of friction, including static and kinetic friction
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lyn83716
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Homework Statement



Block A is on top of Block B, both are of equal mass and size. Block B is being pulled by a hook at a constant velocity. There is friction between the table and block B.

Homework Equations


[/B]
What will each FBD look like? Is there friction between Block A and Block B?

The Attempt at a Solution



What are the downwards forces acting on the bottom block (B)? Is it gravity as well as the normal force of the top block? Do those two added = the normal force of Block B?

Block A FBD:
-force of gravity of block a (Fga) pointing downwards which equals to the force of normal of block a pointing upwards(Fgn)
-friction between block a and block b? i don't know

Block B FBD:
-force of gravity pointing downwards, force of A on B pointing downwards
-normal force of b pointing upwards
-applied force pointing fwd
-force of kinetic friction pointing backwards
 
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lyn83716 said:
What will each FBD look like?
You need to post your own attempt first, as per forum rules.
 
Hi lyn83716,

Welcome to Physics Forums.

Your post is all questions but with no attempt at solution(s) or thoughts on what you think will happen. Can you perhaps try to draw what you think the FBDs for the blocks will look like and show us so that helpers can critique the result (tell you where you're right and what might be incorrect)?
 
lyn83716 said:

Homework Statement



Block A is on top of Block B, both are of equal mass and size. Block B is being pulled by a hook at a constant velocity. There is friction between the table and block B.

Homework Equations


[/B]
What will each FBD look like? Is there friction between Block A and Block B?

The Attempt at a Solution



What are the downwards forces acting on the bottom block (B)? Is it gravity as well as the normal force of the top block? Do those two added = the normal force of Block B?

Attempt:

Block A FBD:
-force of gravity of block a (Fga) pointing downwards which equals to the force of normal of block a pointing upwards(Fgn)
-friction between block a and block b? i don't know

Block B FBD:
-force of gravity pointing downwards, force of A on B pointing downwards
-normal force of b pointing upwards
-applied force pointing fwd
-force of kinetic friction pointing backwards
 
haruspex said:
You need to post your own attempt first, as per forum rules.
I posted my own attempt in the replies
 
gneill said:
Hi lyn83716,

Welcome to Physics Forums.

Your post is all questions but with no attempt at solution(s) or thoughts on what you think will happen. Can you perhaps try to draw what you think the FBDs for the blocks will look like and show us so that helpers can critique the result (tell you where you're right and what might be incorrect)?

I posted the attempt in the reply
 
lyn83716 said:
-friction between block a and block b? i don't know
Are blocks a and b moving with respect to each other, or is block a moving along with block b? Is the system accelerating?
 
lyn83716 said:
Attempt:

Block A FBD:
-force of gravity of block a (Fga) pointing downwards which equals to the force of normal of block a pointing upwards(Fgn)
-friction between block a and block b? i don't know

Block B FBD:
-force of gravity pointing downwards, force of A on B pointing downwards
-normal force of b pointing upwards
-applied force pointing fwd
-force of kinetic friction pointing backwards
Next step is to turn those into equations of the form ΣF=ma.
 

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