System with Pulleys and a Rope That Might Break

[rashida564]

Homework Statement

A system of two fixed pulleys, each of mass M and radius R, is set up as in the
diagram below. The masses and pulleys are connected by light inextensible strings,
which do not slip on the pulleys. A force of magnitude F is applied as shown to
one of the strings, in order to lift the block of mass m. Calculate the maximum
possible acceleration of the block, given that any string will break if the tension in
it exceeds 5Mg. You may assume that the relevant moment of inertia of each pulley
is MR2/2.

Relevant Equations

F=ma
t=Iα

I tried to solve The system of torque as following.
I labelled pulley one is the one in the left, and two as the one on the right.
α=a/R
t1=(-F+T1) R=MR^/2α. So -F+T=Ma/2
for pulley 2 (T1-T2)R=MR^2/2. So T1-T2=Ma/2
for The box T2-mg=ma
add the last two equation, to get that T1-mg=Ma/2+ma.
Multiply by negative for the first equation and add it to get that.
a=(F-mg)/m.
I feel it's wrong since it doesn't involve the mass of the pulley. Or is it the right and I just over complicate the things

 

[kuruman]

It looks like you have adopted the convention that clockwise torques are negative and counterclockwise torques are positive. That's fine, but it should also apply to angular accelerations. Note that the pulleys have opposite angular accelerations but in your equations they are both positive. (The angular acceleration of pulley 2 is missing but it is assumed positive.) Which one of the two should be positive? Also, please use parentheses if you are not using LaTeX. MR^2/2α is interpreted as $\frac{MR^2}{2\alpha}$.

 

[rashida564]

Thanks but how can I know that they have opposite angular acceleration is it always the case in any two pulley problem

 

[rashida564]

I think the second pulley should be the one with positive, angular acceleration. since it's moving counter clock- wise and the first one is negative it's moving clockwise

 

[kuruman]

I think the second pulley should be the one with positive, angular acceleration. since it's moving counter clock- wise and the first one is negative it's moving clockwise

Yes. I think you can finish the problem now.

 

[haruspex]

Thanks but how can I know that they have opposite angular acceleration is it always the case in any two pulley problem

As I mentioned in your other pulley thread, you don't need to predict that correctly. Just be clear which way is positive for each variable and be consistent in the equations.

 

[kuruman]

As I mentioned in your other pulley thread, you don't need to predict that correctly. Just be clear which way is positive for each variable and be consistent in the equations.

I have not seen the other pulley thread. However, isn't it true (in this pulley thread) that the left end of the string must accelerate up if there is any chance of reaching the string's breaking point? If so, then one can predict the sense of rotation of the pulleys.

 

[rashida564]

As I mentioned in your other pulley thread, you don't need to predict that correctly. Just be clear which way is positive for each variable and be consistent in the equations.

If I putted for both of them as moving upward so a is positive for both of them won't that give different answer. because once I putted that both of them will have the same acceleration I got different answer

 

[haruspex]

I have not seen the other pulley thread. However, isn't it true (in this pulley thread) that the left end of the string must accelerate up if there is any chance of reaching the string's breaking point? If so, then one can predict the sense of rotation of the pulleys.

I committed the sin of not reading the actual question in this thread. My remark breaks down when there are nonlinear conditions involved, such as a maximum magnitude, or friction, or the fact that a string cannot support compression.
It can be necessary, after developing the general equations, which may involve modulus signs, to break it into cases. As you write, in the present case it is kind of obvious that the string must accelerate up on the left.

 

[rashida564]

I am confused to which one of them put the maximum value of the tension.

 

[kuruman]

I am confused to which one of them put the maximum value of the tension.

Did you solve the 3 equations to get the tensions and the acceleration?

 

[rashida564]

I tried to start from the beginning I putted the positive direction as the direction in which the pulley will move.
So for the first pulley the one in the left. the rope will move clock wise so I putted the clock wise as positive.
(F-T1)R=(1/2) MR^2(a/R)
So F-ta= (1/2) Ma
For the second pulley the rope will move clockwise
So (T1-T2)R=1/2 MR^2 (a/r)
so T1-T2=1/2 Ma
Last equation for the box
T2-mg=ma
Now we know that maximum tension is 5Mg since T1 and T2 is positive and
T1=1/2Ma+T2
Then T1 is greater than T2 .
Based on that T1=5Mg maximum value
After that I went to put it into the last equation
T2-mg=ma
where T2=5Mg-1/2 Ma
So lastly I found the value of a.
a=(5Mg - mg)/(m+1/5 M)

 

[rashida564]

I hope that I did it It's easier than I expected

 

[kuruman]

You still didn't get it. First of the string (at least the parts that are not wrapped around a pulley) move in straight lines. All points have the same magnitude of linear acceleration and that allows you to say $a=\alpha R$ to relate the angular acceleration to the linear acceleration. Second, one pulley spins faster and faster clockwise and the other faster and faster counterclockwise. This means that if you choose $\alpha_1$ positive, $\alpha_2$ must be negative. The torque equations are $\tau_1=I_1\alpha_1$ and $\tau_2=I_2\alpha_2$. If you set $\alpha_1=a/R$, what should $\alpha_2$ be in terms of the magnitude of the linear acceleration $a$? Third, since you adopted the convention that clockwise is positive, any product $T_iR$ that would produce a clockwise angular acceleration of a particular pulley should appear with a $+$ sign on the left side of the equation. Likewise, any product $T_iR$ that would produce a counterclockwise angular acceleration of a particular pulley should appear with a $-$ sign on the left side of the equation. You need to be consistent as @haruspex advised in post #6.

 

[rashida564]

@kuruman Sorry mate for this stupid question is there any physics behind why the pulley, spend in the opposite directions. What if there's three pulley system how can we know that which two will spend in opposite direction

 

[rashida564]

Do you recommend any textbook to study those idea from, like more complex pulley system. Most pulley system that I saw online are simple without moment on inertia

 

[rashida564]

@kuruman when I followed the convention of clockwise and anti clock wise I got the same equation as the one in my last post
(F-T1)R=1/2 MR^2 (a/R)
So F-T1=1/2 Ma
and (T2-T1)R=-1/2 MR^2 (a/2)
So T1-T2=1/2 M(a)

 

[kuruman]

@kuruman Sorry mate for this stupid question is there any physics behind why the pulley, spend in the opposite directions. What if there's three pulley system how can we know that which two will spend in opposite direction

It's not a stupid question. The correct approach will allow you to answer it on your own. Play a movie in your head and imagine a hand pulling the free end of the string with constant force $F$. What happens to the three straight pieces of the string? The left part goes up and the middle part goes down. If that is the case, how is the left pulley going to turn, clockwise or counterclockwise? Now look at the pulley on the right. The middle part of the string goes down and the right part goes up. If that is the case, how is the right pulley going to turn, clockwise or counterclockwise? It's as simple as that. When you do physics problems, it often helps to play a movie in your head and imagine how a system might evolve in time.

Do you recommend any textbook to study those idea from, like more complex pulley system. Most pulley system that I saw online are simple without moment on inertia

I don't know of any textbooks. "On-line" also includes Physics Forums. Have you tried an internal search? Pulley problems are quite common here. Start with the list of related threads at the bottom of the screen and then branch out to the related threads of the related threads and so on.

 

[rashida564]

Thanks man, so the pulley will rotate the same way the rope will "rotate" Like if I focused on a point on the left pulley it will rotate in clockwise direction so the pulley will also rotate in clockwise direction. Am I right
Sorry for asking lots of questions, for the right have alpha=a/r are we counting r as the radius of the circle even if we applied the force somewhere inside the pulley like in a wheel axle system

 

[kuruman]

@kuruman when I followed the convention of clockwise and anti clock wise I got the same equation as the one in my last post
(F-T1)R=1/2 MR^2 (a/R)
So F-T1=1/2 Ma
and (T2-T1)R=-1/2 MR^2 (a/2)
So T1-T2=1/2 M(a)

The two equations are correct. There is a third equation that involves the hanging mass. That will give you three equations and three unknowns.

 

[rashida564]

The last equation I think it's written up

Last equation for the box
T2-mg=ma
Now we know that maximum tension is 5Mg since T1 and T2 is positive and
T1=1/2Ma+T2
Then T1 is greater than T2 .
Based on that T1=5Mg maximum value
After that I went to put it into the last equation
T2-mg=ma
where T2=5Mg-1/2 Ma
So lastly I found the value of a.
a=(5Mg - mg)/(m+1/5 M)

 

[kuruman]

I agree with your 3 equations and that T1 is the one that reaches the value 5Mg first. I disagree with your final expression for the acceleration when T1=5Mg. Reheck your algebra or post it here if you want it to be looked at.

On edit: The factor $(5Mg-mg)$ up front is correct. It says that if you hang a weight equal to the breaking limit of the string, it will break in the static case of zero acceleration. It's what multiplies that factor that I disagree with.

 

[rashida564]

Thanks mate, I solved for a again I think I made an algebra mistake the answer should be (mg-5Mg)/(m-1/2M)
I hope I didn't make another algebra mistake here.
I got question about angular acceleration alpha isn't always a/R. where R is the radius of the disk or will it depend on the location of application of the force. Or should I ask this question in another post

 

[kuruman]

Thanks mate, I solved for a again I think I made an algebra mistake the answer should be (mg-5Mg)/(m-1/2M)
I hope I didn't make another algebra mistake here.
I got question about angular acceleration alpha isn't always a/R. where R is the radius of the disk or will it depend on the location of application of the force. Or should I ask this question in another post

It's still not right. If you posted your workings, I would be able to point out where you went wrong. You have negative sign problems that you will need to trace.
As for your other question, you can ask here because it is related. Imagine a disk of radius $R$ and moment of inertia $I$ at the rim of which force $F$ is applied. The angular acceleration is $\alpha=\frac{FR}{I}$. This angular acceleration is common to all the points on the disk. The linear acceleration of a point on the rim ($r=R$) is $a_{rim}=\alpha r=\frac{FR}{I}R=\frac{FR^2}{I}$. The linear acceleration of a point halfway from the center to the rim ($r=R/2$) is $a_{halfway}=\alpha r=\frac{FR}{I}(R/2)=\frac{FR^2}{2I}$. It's the distance of the point of interest to the center that counts not necessarily the radius of the disk. Translated from mathematese into English, $a=\alpha r$ says that "The linear acceleration $a$ of a point on a rotating object is equal to the angular acceleration $\alpha$ multiplied by the distance $r$ from the center (or axis) of rotation to the point." Note that this is true regardless of the shape of the object; it could be a disk or a banana. It makes great sense because all points on the rotating object must make one revolution in the same amount of time. Points twice as far from the axis must have twice the acceleration to accomplish this.

 

[rashida564]

Thanks for being patient with me.
First step T1=5Mg So
T2=T1-1/2Ma=M(5g-1/2a)
From last equation
M(5g-1/2 a ) - mg = ma So
5Mg-mg=a(m+1/2 M)
a= (5Mg-mg)/(m + 1/2 M).
I need to be super careful in the exam about the sings.
Mate can you have a look of a similar problem that caused me confusion about angular and tangential acceleration.

 

[kuruman]

a= (5Mg-mg)/(m + 1/2 M).

This matches my answer.

Mate can you have a look of a similar problem that caused me confusion about angular and tangential acceleration.

Not in the next 24 hours. Just post it on a separate thread and someone is very likely to pick it up.

 

[rashida564]

Thanks @kuruman very much for your help. Now I can understand the sign convention as well as solving pulley systems.