Static friction between 2 blocks and tension of string

[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).

 

[goldfish9776]

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).

I can finally understand now . Thank you very much !