Static friction between 2 blocks and tension of string

[goldfish9776]

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

 

[haruspex]

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

 

[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 ?

 

[haruspex]

F= umg

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

 

[goldfish9776]

The other one is static friction force

 

[goldfish9776]

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

 

[haruspex]

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?

 

[haruspex]

The other one is static friction force

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

 

[goldfish9776]

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

Tension of string

 

[haruspex]

Tension of string

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

 

[goldfish9776]

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

 

[SammyS]

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:

 

[goldfish9776]

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

 

[haruspex]

static friction = tension of string when the object is stationary

Right. Does that answer your question in the OP?

 

[goldfish9776]

No, I still don't know why the tension of string isn't exist for block q ...

 

[haruspex]

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?

 

[goldfish9776]

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 ??

 

[goldfish9776]

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)

 

[haruspex]

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.

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.

 

[SammyS]

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.

 

[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...

 

[haruspex]

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.

 

[haruspex]

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.

 

[goldfish9776]

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?

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

 

[haruspex]

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.

 

[goldfish9776]

For mass p , f1=umg=t
For mass q , f2=(mass of p+ q )(u )(g) = t
But tension are nt the same... Why?

 

[haruspex]

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?

 

[goldfish9776]

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.

 

[haruspex]

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.

 

[goldfish9776]

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 ...

 

[haruspex]

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?

 

[goldfish9776]

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]

 

[haruspex]

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.

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.

 

[SammyS]

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 ?
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:

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.

 

[haruspex]

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).