Effect of friction on the tension acting on the string

[decentfellow]

Homework Statement

A weightless string passes through a slit over a pulley. The slit offers frictional force '##f##' to the string. the string carries two weights having masses ##m_1## and ##m_2## where ##m_2 \gt m_1##, then find the acceleration of the weights.

Homework Equations

$$\vec{F}=m\vec{a}$$

The Attempt at a Solution

Now, I have no idea how to deal with the friction with the slit. If the slit provides tension, then should there be a different tension in the string along two different ends of the slit? Just can't think of any other way to incorporate friction in the answer if not that.

 

[jbriggs444]

Now, I have no idea how to deal with the friction with the slit. If the slit provides friction, then should there be a different friction in the string along two different ends of the slit?

I think you meant to write a "different tension" in the string at the two ends of the slit. If so, you are correct.

Of course, as is generally the case with friction, if the friction is adequate to prevent slippage, it exerts only enough force to do so, no more.

 

[decentfellow]

I think you meant to write a "different tension" in the string at the two ends of the slit. If so, you are correct.

Of course, as is generally the case with friction, if the friction is adequate to prevent slippage, it exerts only enough force to do so, no more.

Oops, that was my intention, anyhow edited it.

 

[haruspex]

should there be a different tension in the string along two different ends of the slit?

Yes. Proceed on that basis.

 

[decentfellow]

Yes. Proceed on that basis.

But what should I think while trying to relate the two different tensions at either side of the slit.

 

[conscience]

But what should I think while trying to relate the two different tensions at either side of the slit.

Write ΣF = Ma for the massless string between the slit .This will give you the relation between the two tensions acting at its ends . Thereafter do the usual force balance on the two blocks.

 

[decentfellow]

Write ΣF = Ma for the massless string between the slit .This will give you the relation between the two tensions acting at its ends . Thereafter do the usual force balance on the two blocks.

So after reading your post what I arrived at was that if the (according to the figure accompanying the question) the tension in the string below the slit is ##T_2## and above the slit it is ##T_1##, then we get the relation ##T_2=T_1+f##. Am I correct? If what I have done is indeed correct then what I don't understand is that why is it that when we move up the slit the tension decreases so as to overcome the force of friction, wouldn't the acceleration of the part above the slit have a different acceleration than the one below the slit, and now as the string is inextensible then how would this happen? I know that the string is massless but I can't but i can't think of anything that could clear this doubt of mine, your assistance is very much needed. Ca n you also help me in this @haruspex

 

[haruspex]

So after reading your post what I arrived at was that if the (according to the figure accompanying the question) the tension in the string below the slit is ##T_2## and above the slit it is ##T_1##, then we get the relation ##T_2=T_1+f##. Am I correct? If what I have done is indeed correct then what I don't understand is that why is it that when we move up the slit the tension decreases so as to overcome the force of friction, wouldn't the acceleration of the part above the slit have a different acceleration than the one below the slit, and now as the string is inextensible then how would this happen? I know that the string is massless but I can't but i can't think of anything that could clear this doubt of mine, your assistance is very much needed. Ca n you also help me in this @haruspex

Since the string is massless, the net force on any given portion of the string must be zero. Its acceleration is entirely controlled by the accelerations of the objects attached to it.

 

[decentfellow]

Since the string is massless, the net force on any given portion of the string must be zero. Its acceleration is entirely controlled by the accelerations of the objects attached to it.

Gotta say you always go ahead clear the doubts in the least possible words, just didn't think of the acceleration of the masses attached to the string, and they are indeed same, i.e ##\dfrac{(m_2-m_1)g-f}{m_1+m_2}##.

 

[haruspex]

Gotta say you always go ahead clear the doubts in the least possible words, just didn't think of the acceleration of the masses attached to the string, and they are indeed same, i.e ##\dfrac{(m_2-m_1)g-f}{m_1+m_2}##.

Looks good. Interestingly, it still holds if the system is static.

 

[jbriggs444]

Looks good. Interestingly, it still holds if the system is static.

Careful. The system can be static over a range of nearly-equal masses where the equation predicts a non-zero acceleration.

 

[haruspex]

Careful. The system can be static over a range of nearly-equal masses where the equation predicts a non-zero acceleration.

It says the frictional force is f. Nothing is said about coefficients of friction. If it is static, the frictional force will be whatever makes it static.

 

[decentfellow]

Looks good. Interestingly, it still holds if the system is static.

What does "static system" mean here. If it is the same as the system having no acceleration then howcome the expression that I have written holds even when the system is static wouldn't the acceleration of the system be zero.

 

[haruspex]

What does "static system" mean here. If it is the same as the system having no acceleration then howcome the expression that I have written holds even when the system is static wouldn't the acceleration of the system be zero.

Yes. What does that tell you about f in that case?

 

[decentfellow]

Yes. What does that tell you about f in that case?

Okay then that means that when the system is static then also the friction is acting and is equal to ##(m_2-m_1)g##, and if the friction acting b/w the slit and the string is zero then we have an acceleration ##\dfrac{(m_2-m_1)g}{m_1+m_2}##, but how is that possible because in that case there would be relative motion b/w the slit and the string so friction must be acting b/w hte surface of slit and thread.

 

[haruspex]

when the system is static then also the friction is acting and is equal to (m2−m1)g,

yes.

if the friction acting b/w the slit and the string is zero then we have an acceleration (m2−m1)g/(m1+m2)

Yes.

in that case there would be relative motion b/w the slit and the string so friction must be acting b/w hte surface of slit and thread.

No. If there is no friction there is no friction. There doesn't suddenly become friction because there is movement.
To put it another way, the coefficient of static friction could be equal to that of kinetic friction but it can never be less.

 

[decentfellow]

No. If there is no friction there is no friction. There doesn't suddenly become friction because there is movement.
To put it another way, the coefficient of static friction could be equal to that of kinetic friction but it can never be less.

Ohh! I see what I did wrong there I assumed that the slit doesn't provide any friction whatsoever, i.e. ##\mu = 0##, hence there would be no force of friction.

 

[jbriggs444]

It says the frictional force is f. Nothing is said about coefficients of friction. If it is static, the frictional force will be whatever makes it static.

It does not say that the frictional force is f. It says that the frictional force "offerered" is f. I take that to mean that the f denotes the maximum force that could be achieved by friction, not the actual force. Of course, that is purely an interpretation and opinions can vary.

 

[haruspex]

It does not say that the frictional force is f. It says that the frictional force "offerered" is f. I take that to mean that the f denotes the maximum force that could be achieved by friction, not the actual force. Of course, that is purely an interpretation and opinions can vary.

Yes, that is far from clear. But if it means the maximum frictional force, that would normally imply static. That would make the question unanswerable.