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Another Friction Problem

  • Thread starter Peach
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1. Homework Statement
Block A in the figure View Figure weighs w_1 and block B weighs w_2. The coefficient of kinetic friction between all surfaces is u_k.

Find the magnitude of the horizontal force \vec{F} necessary to drag block B to the left at constant speed if A is held at rest (figure (b)).


2. Homework Equations
F = ma
a = 0

3. The Attempt at a Solution
I attempted by drawing two FBDs for A and B. I have for

B: w_2 downward, F_aonb downward, normal force upward, F to the left dragging, and fk_ground
A: w_1 downward, normal force upward, F_bona, and fk_ab

Then I solved for the normal force and I got n = (w_1+w_2)/2 and then I plugged it into the fk = ukn, but well wrong answer.

I have a few questions because I don't really understand the whole picture. There's two frictional forces acting on B right? From A and from the ground? Is this correct? Also, the normal force for both B and A...are they the same or are they different? They're different right?
 

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PhanthomJay
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Homework Helper
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1. Homework Statement
Block A in the figure View Figure weighs w_1 and block B weighs w_2. The coefficient of kinetic friction between all surfaces is u_k.

Find the magnitude of the horizontal force \vec{F} necessary to drag block B to the left at constant speed if A is held at rest (figure (b)).


2. Homework Equations
F = ma
a = 0

3. The Attempt at a Solution
I attempted by drawing two FBDs for A and B. I have for

B: w_2 downward, F_aonb downward, normal force upward, F to the left dragging, and fk_ground what about the magnitude and direction of the friction force from A on B ?
A: w_1 downward, normal force upward, F_bona, and fk_ab Specify the directions of the ones you have identified.

Then I solved for the normal force and I got n = (w_1+w_2)/2 and then I plugged it into the fk = ukn, but well wrong answer.how did you get this?

I have a few questions because I don't really understand the whole picture. There's two frictional forces acting on B right? From A and from the ground? Is this correct? correct Also, the normal force for both B and A...are they the same or are they different? They're different right?correct[color]
Draw the FBD's again and identify the forces on each block. The normal force on block A is the force upwards of B on A.
 
Doc Al
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B: w_2 downward, F_aonb downward, normal force upward, F to the left dragging, and fk_ground
You left out the friction force of A on B. Note also: the force you call F_aonb downward is the normal force between A and B. A and B exert two forces on each other: a normal force and a friction force.
A: w_1 downward, normal force upward, F_bona, and fk_ab
Seems like you are counting the normal force twice: F_bona is the normal force. And you left out the tension in the rope pulling on A. (You don't need to analyze the forces on A to solve this problem, but it's good exercise.)

Then I solved for the normal force and I got n = (w_1+w_2)/2 and then I plugged it into the fk = ukn, but well wrong answer.
I assume you mean the normal force between the ground and B. How did you solve for it? (Consider vertical forces only.)

I have a few questions because I don't really understand the whole picture. There's two frictional forces acting on B right? From A and from the ground? Is this correct?
Absolutely correct.
Also, the normal force for both B and A...are they the same or are they different? They're different right?
Yes. The normal force between A & B is different from the normal force between B & ground.
 
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Okay, the friction force from A to B is going in the same direction as the friction force from the ground to B right? Because friction forces are always in the opposite direction from the acceleration and perpendicular to normal forces..? If that's not wrong, then this is what I'm getting:

n_gb: n_ab + w_2

F: fk_ab + fk_gb

Is this correct? Am I going in the right direction, no pun intended?

Edit: Oh! And n_ab is w_1 right? Because the normal force on A = n_bona = w_1? That's what I'm getting from my FBD of A...
 
Last edited:
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,120
463
Okay, the friction force from A to B is going in the same direction as the friction force from the ground to B right? Because friction forces are always in the opposite direction from the acceleration and perpendicular to normal forces..? If that's not wrong, then this is what I'm getting:

n_gb: n_ab + w_2

F: fk_ab + fk_gb

Is this correct? Am I going in the right direction, no pun intended?

Edit: Oh! And n_ab is w_1 right? Because the normal force on A = n_bona = w_1? That's what I'm getting from my FBD of A...
Yes, this seems correct....
 
80
0
I got the answer. Many thanks for both the help. ^^
 

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