Static friction between 2 blocks and tension of string

In summary: If block Q moves to the right, block P will move to the right, and the tension on Q will be the same as the tension on P.
  • #1
goldfish9776
310
1

Homework Statement



the maximum force is F=F1 +F2 + T as shown in the textbook...
I can't 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) ??

Homework Equations

The Attempt at a Solution

 

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  • #2
It says string, not spring.
Consider the horizontal forces on block P. Given equilibrium, what equation can you write?
 
  • #3
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
goldfish9776 said:
F= umg
That's only one of the two horizontal forces acting on P. What's the other one?
 
  • #5
The other one is static friction force
 
  • #6
How about q ? There's only static friction force? No tension of string acted on q ?
 
  • #7
goldfish9776 said:
How about q ? There's only static friction force? No tension of string acted on q ?
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
goldfish9776 said:
The other one is static friction force
No, ##F=\mu mg## is the static frictional force. What is the other horizontal force on P?
 
  • #9
haruspex said:
No, ##F=\mu mg## is the static frictional force. What is the other horizontal force on P?
Tension of string
 
  • #10
goldfish9776 said:
Tension of string
Right. If there's static equilibrium, what does that tell you about those two forces?
 
  • #11
The object is stationary until the applied force larger than the static frictional force, the object starts to move
 
  • #12
goldfish9776 said:
The object is stationary until the applied force larger than the static frictional force, the object starts to move
That almost answers haruspex's question.

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:

upload_2015-6-26_10-6-54.png
 
  • #13
SammyS said:
That almost answers haruspex's question.

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:

View attachment 85250
static friction = tension of string when the object is stationary
 
  • #14
goldfish9776 said:
static friction = tension of string when the object is stationary
Right. Does that answer your question in the OP?
 
  • #15
No, I still don't know why the tension of string isn't exist for block q ...
 
  • #16
goldfish9776 said:
No, I still don't know why the tension of string isn't exist for block q ...
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
haruspex said:
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?
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 isn't equal to F2[ (u )(mass of p + mass of q)(g) ] only ??
 
  • #18
haruspex said:
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?
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
goldfish9776 said:
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)
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.
goldfish9776 said:
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 isn't equal to F2[ (u )(mass of p + mass of q)(g) ] only ??
If block Q moves to the right, block P will move to the left. It's an inextensible string, not a spring.
 
  • #20
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
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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  • #22
goldfish9776 said:
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...
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
goldfish9776 said:
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...
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
haruspex said:
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.
Something like this?
uploadfromtaptalk1435502046398.jpg

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
 
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  • #25
goldfish9776 said:
Something like this? View attachment 85310
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 ?
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.
 
  • #26
For mass p , f1=umg=t
For mass q , f2=(mass of p+ q )(u )(g) = t
But tension are nt the same... Why?
 
  • #27
goldfish9776 said:
For mass q , f2=(mass of p+ q )(u )(g) = t
That's not right.
Which way does F2 act on Q? Which way does T act on Q? What other force acts on Q?
 
  • #28
Which way does F2 act on Q?
F2 is directed to the left , as shown in the figure

Which way does T act on Q?
Tension of string is also directed to the left

What other force acts on Q?
There's F1 , F2 , and T act on Q.
 
  • #29
goldfish9776 said:
Which way does F2 act on Q?
F2 is directed to the left , as shown in the figure

Which way does T act on Q?
Tension of string is also directed to the left

What other force acts on Q?
There's F1 , F2 , and T act on Q.
There are four horizontal forces in the picture that act on Q. All four need to be in the equation.
 
  • #30
haruspex said:
There are four horizontal forces in the picture that act on Q. All four need to be in the equation.
ok, F1 + F2 + T = F...
I still don't understand why the tension is calculated in this way ...
 
  • #31
goldfish9776 said:
ok, F1 + F2 + T = F...
I still don't understand why the tension is calculated in this way ...
In post #26 you stated, correctly, that T=F1=umg, where T is the tension in the upper part of the string.
As I noted in post #25, the tension in the lower part of the string must also be T, or the pulley would rotate.
Which of those two steps do you not understand?
 
  • #32
haruspex said:
In post #26 you stated, correctly, that T=F1=umg, where T is the tension in the upper part of the string.
As I noted in post #25, the tension in the lower part of the string must also be T, or the pulley would rotate.
Which of those two steps do you not understand?

I don't understand why the tension of string is 0.981N . I knew that the tension of the upper and lower part of string must be the same , in order for the pulley system to be functional . why can't it be 1.373N ??
As we all know , F1= frictional static force between P and Q..Why can't F2 equal to tension of string ?[/QUOTE]
 
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  • #33
goldfish9776 said:
I don't understand why the tension of string is 0.981N
We've been around this already. The static balance of horizontal forces on P leads to this.
goldfish9776 said:
Why can't F2 equal to tension of string ?
Why should T=F2? What equation leads to that? The static balance of horizontal forces on Q has four forces in it, and you wrote it correctly in post #30.
If you think T should be F2 then you need to explain why you think that.
 
  • #34
goldfish9776 said:
Something like this? View attachment 85310
uploadfromtaptalk1435502046398-jpg.85310.jpg

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
Compare the above with the figure from your book:
upload_2015-6-26_10-6-54-png.85250.png


The vectors labeled T and T correspond to the pull of the string on each block. This is the string you have which passes over the pulley.

The vector labeled F, going to the right from block Q, represents some external force. This is the force necessary to move Q at a constant velocity, v. The portion of the problem we can see does not state how this force is applied to Q; whether by some string or by other means.

In your figure, block Q is being pulled to the right by some force which you have not included.
 
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  • #35
goldfish9776 said:
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
I'm fishing around trying to understand what wrong idea is blocking you on this. Maybe the clue is in the above two statements.
"You" do not pull on either of those parts of the string. "You" supply the force F going off to the right.
Without friction, block P would move to the left and block Q would move to the right. On block Q, T and F2 act in the same direction, so they add.
Your notion that the tension would somehow match F2 could only be right if Q were tending to slide left, making T and F2 oppose (and if P and F were not present).
 
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<h2>1. What is static friction?</h2><p>Static friction is a force that prevents two surfaces from sliding against each other when they are in contact and at rest.</p><h2>2. How is static friction different from kinetic friction?</h2><p>Static friction occurs when two surfaces are at rest, while kinetic friction occurs when two surfaces are in motion.</p><h2>3. What factors affect the magnitude of static friction?</h2><p>The magnitude of static friction depends on the nature of the surfaces in contact, the normal force between the surfaces, and any additional external forces acting on the objects.</p><h2>4. How is tension in a string related to static friction between two blocks?</h2><p>The tension in a string is equal to the magnitude of the static friction force between two blocks. This is because the string is what transmits the force between the two blocks, and the force of static friction acts in the opposite direction to the tension in the string.</p><h2>5. Can the magnitude of static friction ever be greater than the applied force?</h2><p>No, the magnitude of static friction can never be greater than the applied force. If the applied force is greater than the maximum static friction force, the objects will start to move and kinetic friction will take over.</p>

FAQ: Static friction between 2 blocks and tension of string

1. What is static friction?

Static friction is a force that prevents two surfaces from sliding against each other when they are in contact and at rest.

2. How is static friction different from kinetic friction?

Static friction occurs when two surfaces are at rest, while kinetic friction occurs when two surfaces are in motion.

3. What factors affect the magnitude of static friction?

The magnitude of static friction depends on the nature of the surfaces in contact, the normal force between the surfaces, and any additional external forces acting on the objects.

4. How is tension in a string related to static friction between two blocks?

The tension in a string is equal to the magnitude of the static friction force between two blocks. This is because the string is what transmits the force between the two blocks, and the force of static friction acts in the opposite direction to the tension in the string.

5. Can the magnitude of static friction ever be greater than the applied force?

No, the magnitude of static friction can never be greater than the applied force. If the applied force is greater than the maximum static friction force, the objects will start to move and kinetic friction will take over.

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