# Homework Help: Static friction between 2 blocks and tension of string

1. Jun 26, 2015

### goldfish9776

1. The problem statement, all variables and given/known data

the maximum force is F=F1 +F2 + T as shown in the textbook....
I cant understand why the tension of spring is also = F1...Is it because of the writer assume the spring is directly attached to the part P (upper part of the block) ??
2. Relevant equations

3. The attempt at a solution

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• ###### IMG_20150626_145554[1].jpg
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2. Jun 26, 2015

### haruspex

It says string, not spring.
Consider the horizontal forces on block P. Given equilibrium, what equation can you write?

3. Jun 26, 2015

### goldfish9776

F= umg
Is it because of the string is attached to the block p ? So we should only consider the tension acted on block p ?

4. Jun 26, 2015

### haruspex

That's only one of the two horizontal forces acting on P. What's the other one?

5. Jun 26, 2015

### goldfish9776

The other one is static friction force

6. Jun 26, 2015

### goldfish9776

How about q ? There's only static friction force? No tension of string acted on q ?

7. Jun 26, 2015

### haruspex

I'm trying to lead you to the answer to your question. Stick with P. There are two horizontal forces on P. Look at the diagram. What are they?

8. Jun 26, 2015

### haruspex

No, $F=\mu mg$ is the static frictional force. What is the other horizontal force on P?

9. Jun 26, 2015

### goldfish9776

Tension of string

10. Jun 26, 2015

### haruspex

Right. If there's static equilibrium, what does that tell you about those two forces?

11. Jun 26, 2015

### goldfish9776

The object is stationary until the applied force larger than the static frictional force, the object starts to move

12. Jun 26, 2015

### SammyS

Staff Emeritus

How do the two forces on P (static friction and tension of string) compare when the objects are stationary?
By the way, here is a portion of the image you provided:

13. Jun 26, 2015

### goldfish9776

static friction = tension of string when the object is stationary

14. Jun 27, 2015

### haruspex

15. Jun 27, 2015

### goldfish9776

No, I still dont know why the tension of string isnt exist for block q ...

16. Jun 27, 2015

### haruspex

You asked why the tension is equal to the frictional force. That is now explained.
What do you mean by the tension of the string "not existing" for Q? Do you mean it does not act on Q? Clearly it does. What makes you think it does not?

17. Jun 27, 2015

### goldfish9776

ok , i am confused now . for block Q to start to move( block p will remain stationary on top of the block q when block q starts to move ) , why the force F isnt equal to F2[ (u )(mass of p + mass of q)(g) ] only ??

18. Jun 27, 2015

### goldfish9776

I thought the tension of spring is not existing of Q simply because i saw the max force for Q is T+F1 +F2 , which the tension of string is equal to F1 (only mass of block P involved , but not mass of block Q)

19. Jun 27, 2015

### haruspex

As os clear from the diagram, the horizontal forces acting on Q are af1 (but this is opposite to the F1 acting on P), F2, and T.
From the earlier analysis of block P, T is equal and opposite to the F1 acting on P, and therefore completely equal to the F1 acting on Q.
If block Q moves to the right, block P will move to the left. It's an inextensible string, not a spring.

20. Jun 27, 2015

### SammyS

Staff Emeritus
The force that block Q exerts on block P is equal in magnitude, but opposite in direction to the force that block P exerts on block Q . (Newton's 3rd Law) The magnitude of that force is labeled F1 in the diagram. When acting on P, the force is to the right. When acting on Q the force is to the left.

21. Jun 28, 2015

### goldfish9776

Can someone draw me a diagram so that i can understand better . Since the question say both blocks are connected by a string . So , i assume the blocks are connected in this way...Correct me if i'm wrong...

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• ###### H.png
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22. Jun 28, 2015

### haruspex

That's wrong. Read the question again. It's just one piece of string that goes off to the left from P, around a pulley with a fixed axis, and doubles back to connect to Q. That's why the tension is the same.

23. Jun 28, 2015

### haruspex

That's wrong. Read the question again. It's just one piece of string that goes off to the left from P, around a pulley with a fixed axis, and doubles back to connect to Q. That's why the tension is the same.

24. Jun 28, 2015

### goldfish9776

Something like this?
If this is true , then which part of rope should I pull ? the upper part connected to the block P or the lower part connected to the part Q ? How can be the tension be the same ?
If I pull the upper part , then I agree the tension of the rope should be = F1
If I pull the lower part of the rope , then I think that the tension of the rope should be = F2

Last edited: Jun 28, 2015
25. Jun 28, 2015

### haruspex

Neither part of that string is being pulled directly. Q is being pulled to the right by force F.
If the tensions in the upper and lower parts of the string are not the same then there will be a net torque on the pulley, making it turn.
Consider the forces on each block separately.
Write down the static equilibrium equation for the horizontal forces on P, then do the same for Q.