Does the tension in a string act on the body to which the pulley is attached?

[Hamiltonian]

Homework Statement

in the given figure a block of mass m is placed on a wedge of mass M all surfaces are frictionless and the pulley is massless and light the string can be assumed to be ideal name all the forces acting on the wedge.

Relevant Equations

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The normal reaction from the ground and from block m, the force of gravity are the forces I feel should act on the wedge but since the wedge is a perfectly rigid body and the pulley(which is massless) is attached to it so will tension also act on the wedge as well?

 

[etotheipi]

There will be a force, but it's worth clarifying a few things. A tension force is an axial force, an as such it only really makes sense to talk about the force of tension either acting on another part of the rope, or on a body attached axially to the rope.

When you have a rope in contact with a pulley, really there is a contact force (normal force, and friction force if that is applicable) between each small section of the rope and the pulley. You can, however, deduce that the sum of the contact forces of all of the bits of the rope in contact with the pulley are equivalent to the action of two forces whose magnitudes are the tension $T$, acting along the two respective directions of the rope on either side of the pulley. If there is friction, these two forces will not be equal in magnitude (in fact they are related by a factor $e^{\mu \theta}$), and you also get a resultant torque.

(N.B. you can prove this for yourself via integration, or more simply you can consider a system comprising of the pulley + the section of rope in contact with the pulley, which is just acted upon by two axial tension forces).

 

[Hamiltonian]

When you have a rope in contact with a pulley, really there is a contact force (normal force, and friction force if that is applicable) between the rope and the pulley.

In the given scenario the pulley is frictionless and the string is massless so there can neither be a normal reaction nor friction (hence no contact force)?

You can, however, deduce that the sum of the contact forces of all of the bits of the rope in contact with the pulley are equivalent to the action of two forces whose magnitudes are the tension $T$, acting along the two respective directions of the rope on either side of the pulley. If there is friction, these two forces will not be equal in magnitude (in fact they are related by a factor $e^{\mu \theta}$), and you also get a resultant torque.

how would I prove this in the given example?

 

[etotheipi]

In the given scenario the pulley is frictionless and the string is massless so there can neither be a normal reaction nor friction (hence no contact force)?

If the pulley is frictionless there can be no friction force, but there definitely can be a normal reaction force between each part of the rope and the pulley! Idealised massless bodies can experience normal forces so long as they are in contact!

Consider a string wrapped around a pulley, subtending a total angle $\alpha$. Now look at a tiny piece of the string subtending an angle $d\theta$, which is acted upon by two tension forces of magnitude $T$ acting at $\frac{d\theta}{2}$ below the local horizontal, in addition to a normal contact force $dN$ from the pulley. Equilibrium in the radial direction dictates that$$2T\sin{\frac{d\theta}{2}} \approx Td\theta = dN$$Hence, looking at the resultant $y$ component of force on the pulley,$$dN_y = -T\cos{\theta} d\theta \implies N_y = -T\int_{-\frac{\alpha}{2}}^{\frac{\alpha}{2}} \cos{\theta} d\theta = -2T\sin{\frac{\alpha}{2}}$$i.e. the resultant contact force on the pulley is equivalent to two tangential forces $T$.

Maybe now you want to do the frictional case, in which case you can again consider a small element of string but now you need to account for the fact that $T = T(\theta)$. In the limiting case, a friction force $\mu dN \approx \mu T(\theta) d\theta$ acts on each little segment of string, so the force balance in the tangential direction gives$$T(\theta +d\theta) = \mu T(\theta) d\theta + T(\theta)$$ $$\int \frac{dT(\theta)}{T(\theta)} = \int \mu d\theta$$You can finish that off to prove the exponential relationship! Then you can try and show that the total vertical contact force in this more complicated case with variable tension is likewise equivalent to two tangential forces.

 

[TSny]

a friction force $\mu dN \approx \mu T(\theta) d\theta$ acts on each little segment of string,

This assumes that the segments are on the verge of slipping,

 

[etotheipi]

This assumes that the segments are on the verge of slipping,

You're right; I did try to preface it by saying "in the limiting case". I don't know if a formula exists in the case that friction is not limiting (I could be wrong ), in which case the factor by which tension increases would need to be obtained by equations relating to other parts of the system.

 

[TSny]

You're right; I did try to preface it by saying "in the limiting case".

Yes. I missed that. Sorry.

I don't know if a formula exists in the case that friction is not limiting (I could be wrong ), in which case the factor by which tension increases would need to be obtained by equations relating to other parts of the system.

I don't believe there would be a unique solution for $T(\theta)$ if slipping is not imminent.

 

[Lnewqban]

... The normal reaction from the ground and from block m, the force of gravity are the forces I feel should act on the wedge but since the wedge is a perfectly rigid body and the pulley(which is massless) is attached to it so will tension also act on the wedge as well?

Will the way in which the wedge will react to the non-balanced situation be the same with and without the string?
If not, how would the blind wedge “know” that there is a string in the picture?

 

[haruspex]

how would I prove this in the given example?

Consider the little section of string in contact with the pulley as a free body. It is subject to two tension forces at some angle to each other and a force from the wedge-pulley system. Since it is massless, these must be in balance, so you can deduce the force from the pulley.

 

[JD_PM]

Here's an sketch of what @etotheipi nicely explained at #4.

Notice that if we were to take $T$ as constant (i.e. $T \neq T(\theta)$) it would imply that the pulley would be frictionless.

$$-T\cos(\Delta \theta/2) -f +T\cos(\Delta \theta/2)=0 \ \Rightarrow \ f=0$$

 

[haruspex]

Here's an sketch of what @etotheipi nicely explained at #4.

View attachment 268163
Notice that if we were to take $T$ as constant (i.e. $T \neq T(\theta)$) it would imply that the pulley would be frictionless.

$$-T\cos(\Delta \theta/2) -f +T\cos(\Delta \theta/2)=0 \ \Rightarrow \ f=0$$

Nice picture, but that is not what is meant by a "frictionless pulley". That expression is used to mean that the axle is frictionless, i.e. no frictional torque. Generally we assume there is no slipping between the pulley and the rope. This matters when the pulley may be accelerating and the moment of inertia of the pulley may result in a change in tension.

Oh, and to be even more pedantic, T=T(θ) doesn't rule out the constant case.

 

[JD_PM]

Nice picture, but that is not what is meant by a "frictionless pulley". That expression is used to mean that the axle is frictionless, i.e. no frictional torque. Generally we assume there is no slipping between the pulley and the rope. This matters when the pulley may be accelerating and the moment of inertia of the pulley may result in a change in tension.

Thanks flor the comment haruspex but I am afraid I do not follow your argument. Doesn't 'no frictional torque' mean $f=0$, as I suggested?

Oh, and to be even more pedantic, T=T(θ) doesn't rule out the constant case.

Please notice I wrote $T \neq T(\theta)$

 

[etotheipi]

I think you are both right, but just talking about slightly different things? There can be friction acting at the axle which exerts a resistive couple on the pulley, and there can also be friction at the pulley-rope interface which can likewise exert a torque on the pulley.

N.B. that since the string is massless, the net force on any small piece must still be exactly zero even if the pulley is accelerating, so the same analysis holds as for the static case.

 

[haruspex]

Thanks flor the comment haruspex but I am afraid I do not follow your argument. Doesn't 'no frictional torque' mean $f=0$, as I suggested?

I think you mean, doesn’t f=0 imply no frictional torque?
That is true, but I still feel it was likely to be misinterpreted by a student.

Please notice I wrote $T \neq T(\theta)$

Yes, but my point is that unchanging tension is not expressed by $T \neq T(\theta)$. It is always the case that $T = T(\theta)$, but the function can be constant. As I said, I was being rather pedantic.