Why is Block B not experiencing a normal force in this static friction problem?

In summary, the problem involves three stone blocks with weights of 60lb, 150lb, and 40lb, and a coefficient of static friction of 0.3 between the blocks and 0.5 between the blocks and the floor. The goal is to determine the smallest horizontal force P required to move block C. The assumption that block B slips up leads to the conclusion that there is no normal force on block B from the ground, making the free body diagram of block B different from the standard diagram.
  • #1
OmniNewton
105
5

Homework Statement


How come for the following image when I assume block B slips up, block A does not move. The free body diagram at block B does not have a normal from the ground. This is a static friction problem.
c8159d667350baaca3628c32559c52fb.png
[/B]

Homework Equations


No relevant equations all theory[/B]

The Attempt at a Solution


This should be the Free body diagram but why does B have no normal from the ground when this assumption is made?
9125af555fd21b6e87a783fb1a808f7c.png
 
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  • #2
OmniNewton said:

Homework Statement


How come for the following image when I assume block B slips up, block A does not move. The free body diagram at block B does not have a normal from the ground. This is a static friction problem.
c8159d667350baaca3628c32559c52fb.png
[/B]

Homework Equations


No relevant equations all theory[/B]

The Attempt at a Solution


This should be the Free body diagram but why does B have no normal from the ground when this assumption is made?
9125af555fd21b6e87a783fb1a808f7c.png

It's not clear what the original problem statement was, since you didn't provide it.

When blocks A and C are forced together, block B is going to slide upwards on the sloped face of block C. Once Block B is just a minute distance off the ground, there is no more normal force on the block from that direction.
 
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  • #3
I would assume if block B is slipping up, it is no longer in contact with the ground and therefore there cannot be a normal force.
 
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  • #4
The original problem statement was:
The three stone blocks have weights of 60lb 150lb and 40 lb respectively Determine the smallest horizontal force P that must be applied to block C in order to move this block. The coefficient of static friction between the blocks is 0.3 and between the floor is 0.5
 
  • #5
Based on the question, the only way to move C, is to lift B. Therefore, no normal.
 
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  • #6
rpthomps said:
Based on the question, the only way to move C, is to lift B. Therefore, no normal.
Excellent, thank you that is what I thought. I did the question originally assuming the normal and the system was statically indeterminate I was unsure how to solve until I was told this FBD did not contain a normal but I was unsure as to why
 
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  • #7
rpthomps said:
Based on the question, the only way to move C, is to lift B. Therefore, no normal.
Is that from checking with the actual numbers? With some combinations of weights etc., the three blocks would slide horizontally, no lifting.
 
  • #8
haruspex said:
Is that from checking with the actual numbers? With some combinations of weights etc., the three blocks would slide horizontally, no lifting.

No, I didn't use numbers. I was trying to visualize the block moving without lifting...I couldn't see it.
 
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  • #9
rpthomps said:
No, I didn't use numbers. I was trying to visualize the block moving without lifting...I couldn't see it.
Despite the way the blocks are drawn, block A is much lighter than block B. If it's light enough it will slide instead of B rising. It's easy to visualise - the three blocks slide as if one unit.
Note that the clause "block A does not move" in the OP is not part of the problem specification; it is an observed consequence of the assumption that B slides up.
 
  • #10
haruspex said:
Despite the way the blocks are drawn, block A is much lighter than block B. If it's light enough it will slide instead of B rising. It's easy to visualise - the three blocks slide as if one unit.
Note that the clause "block A does not move" in the OP is not part of the problem specification; it is an observed consequence of the assumption that B slides up.

I guess I just latched onto the phrase Block A does not move. I am not sure if that was a part of the problem originally but that was the assumption I was basing my thoughts on. If it can move, I agree, there is a possibility that the whole unit could move together. In that case, yes I can visualize Block B traveling horizontally. :)
 
  • #11
rpthomps said:
I guess I just latched onto the phrase Block A does not move. I am not sure if that was a part of the problem originally but that was the assumption I was basing my thoughts on. If it can move, I agree, there is a possibility that the whole unit could move together. In that case, yes I can visualize Block B traveling horizontally. :)
The actual problem statement appears to be in post #4.
 

Related to Why is Block B not experiencing a normal force in this static friction problem?

1. What is a free body diagram?

A free body diagram is a visual representation of the forces acting on an object. It shows the magnitude and direction of these forces, and is used to analyze the motion of the object.

2. Why is free body diagram logic important in science?

Free body diagram logic is important because it allows scientists to understand the forces that are acting on an object and how they affect its motion. This is crucial in areas such as physics, engineering, and biomechanics.

3. How do you create a free body diagram?

To create a free body diagram, you first need to identify all the forces acting on the object. These can include gravitational force, normal force, frictional force, and applied forces. Then, draw a simple sketch of the object and label each force with its magnitude and direction.

4. What is the purpose of using free body diagram logic?

The purpose of using free body diagram logic is to simplify complex systems and understand the relationship between forces and motion. It allows scientists to make predictions and analyze the behavior of objects under different conditions.

5. Can free body diagram logic be used for non-rigid bodies?

Yes, free body diagram logic can be used for both rigid and non-rigid bodies. However, the analysis may be more complex for non-rigid bodies due to the additional consideration of deformation and internal forces.

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