Static friction between 2 blocks and tension of string

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The discussion revolves around understanding the forces acting on two blocks connected by a string, focusing on static friction and tension. It clarifies that the tension in the string is equal to the static friction force acting on block P when both blocks are stationary. The confusion arises regarding the forces acting on block Q, where participants debate the role of tension and static friction. It is emphasized that both blocks are influenced by the same tension due to the string's configuration, and the forces must be analyzed separately for each block. Ultimately, the interaction of these forces determines the conditions for motion and equilibrium in the system.
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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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It says string, not spring.
Consider the horizontal forces on block P. Given equilibrium, what equation can you write?
 
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 ?
 
goldfish9776 said:
F= umg
That's only one of the two horizontal forces acting on P. What's the other one?
 
The other one is static friction force
 
How about q ? There's only static friction force? No tension of string acted on q ?
 
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?
 
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?
 
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
 
Last edited:
  • #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 ...
 

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