Pulleys and Springs Homework Help

[Saitama]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

At equilibrium conditions, the forces acting on $m_2$ are the tension due to the strings and its weight. $T=m_2g+T'$ where T is the tension in the string above $m_2$ and T' is the tension in the string below it. Equating forces for $m_1$, $T/2=m_1g \Rightarrow T=2m_1g$. I am not sure what to do next? How would I go about finding the initial acceleration of B?

Any help is appreciated. Thanks!

 

[PhanthomJay]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

At equilibrium conditions, the forces acting on $m_2$ are the tension due to the strings and its weight. $T=m_2g+T'$ where T is the tension in the string above $m_2$ and T' is the tension in the string below it. Equating forces for $m_1$, $T/2=m_1g \Rightarrow T=2m_1g$.

That is correct for the given equilibrium condition.

I am not sure what to do next? How would I go about finding the initial acceleration of B?

Any help is appreciated. Thanks!

Do the free body diagrams again, with the string cut, but this time, after identifying forces, use Newton 2 (the accelerations of the 2 masses may differ).

 

[consciousness]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

At equilibrium conditions, the forces acting on $m_2$ are the tension due to the strings and its weight. $T=m_2g+T'$ where T is the tension in the string above $m_2$ and T' is the tension in the string below it. Equating forces for $m_1$, $T/2=m_1g \Rightarrow T=2m_1g$. I am not sure what to do next? How would I go about finding the initial acceleration of B?

Any help is appreciated. Thanks!

For the FBD of m1 shouldn't it be

$2T=m_1g$

As two strings pass over?

My approach to this question is that due to the spring the Tension in the string(which isn't cut) before and after cutting will remain the same. This is because the elongation of the spring will not change JUST after cutting. Next after making FBD of m2 we can get the answer. But I am not 100% sure.

 

[PhanthomJay]

For the FBD of m1 shouldn't it be

$2T=m_1g$

As two strings pass over?

Oh sure, thanks for the catch, my apologies, I glanced over that too quickly.

My approach to this question is that due to the spring the Tension in the string(which isn't cut) before and after cutting will remain the same. This is because the elongation of the spring will not change JUST after cutting. Next after making FBD of m2 we can get the answer. But I am not 100% sure.

It will have shortened an infinitesimal distance just after the cut in an infinitesimal time, but still have an initial acceleration dv/dt.

 

[Saitama]

Do the free body diagrams again, with the string cut, but this time, after identifying forces, use Newton 2 (the accelerations of the 2 masses may differ).

When the string is cut, the forces acting on $m_2$ are the tension and its weight. For $m_2$, $m_2g-T=m_2a_2$. The forces acting on $m_1$ is the tension and its weight. $m_1g-2T=m_1a_1$. Have I got this right?

For the FBD of m1 shouldn't it be

$2T=m_1g$

As two strings pass over?

Thanks for the catch!

 

[PhanthomJay]

When the string is cut, the forces acting on $m_2$ are the tension and its weight. For $m_2$, $m_2g-T=m_2a_2$. The forces acting on $m_1$ is the tension and its weight. $m_1g-2T=m_1a_1$. Have I got this right?

Yes, provided you have the correct relation and signage between a1 and a2.

 

[Saitama]

Yes, provided you have the correct relation and signage between a1 and a2.

But I am still not sure what to do next? Can I have some hints? :)

 

[PhanthomJay]

But I am still not sure what to do next? Can I have some hints? :)

You have 3 unknowns T, a1, and a2, but so far you have listed just 2 equations. You need another one that describes the relationship between a1 and a2.

a1 and a2 are not the same.

 

[Saitama]

You have 3 unknowns T, a1, and a2, but so far you have listed just 2 equations. You need another one that describes the relationship between a1 and a2.

a1 and a2 are not the same.

$a_1=2a_2$?

 

[PhanthomJay]

$a_1=2a_2$?

Which mass moves the furthest and fastest over an initial time, 1 or 2? Sometimes it is best to draw stop motion sketches and visualize the movement.

 

[Saitama]

Which mass moves the furthest and fastest over an initial time, 1 or 2? Sometimes it is best to draw stop motion sketches and visualize the movement.

I can't figure this out. I don't know what to do with the spring. :(

 

[PhanthomJay]

I can't figure this out. I don't know what to do with the spring. :(

i suppose if you don't know what to do with the spring, try ignoring it and replace it with a cord! The problem asks for the initial acceleration of m2, that is, it's instantaneous acceleration at the instant the lower string is cut. The spring might be very stiff with very little elongation under load, or it might be very stretchy like that wonderful slinky toy, with a lot of elongation under load. But in either case, the spring will compress when the cut is made, but initially, it doesn't matter, what matters is that the mass m2 will rise up, and the mass m1 and the pulley supporting m1 will fall downward. So pretend you have a stiff spring, make it an inextensible cord if you will, and you must then answer the question about the a2 and a1 relationship.

 

[Saitama]

i suppose if you don't know what to do with the spring, try ignoring it and replace it with a cord! The problem asks for the initial acceleration of m2, that is, it's instantaneous acceleration at the instant the lower string is cut. The spring might be very stiff with very little elongation under load, or it might be very stretchy like that wonderful slinky toy, with a lot of elongation under load. But in either case, the spring will compress when the cut is made, but initially, it doesn't matter, what matters is that the mass m2 will rise up, and the mass m1 and the pulley supporting m1 will fall downward. So pretend you have a stiff spring, make it an inextensible cord if you will, and you must then answer the question about the a2 and a1 relationship.

Here's how I look at it. If $m_2$ moves up a distance x, the pulley to which $m_1$ is attached should move 2x. Right?

 

[PhanthomJay]

Here's how I look at it. If $m_2$ moves up a distance x, the pulley to which $m_1$ is attached should move 2x. Right?

If m2 moves up x, then pulley of m1 moves down x/2. Right?

 

[Saitama]

Sorry for such a late reply, I had some problems with my internet connection.

If m2 moves up x, then pulley of m1 moves down x/2. Right?

Ugh..that's the biggest problem I face in these type of problems. If the acceleration of m_2 is a, then acceleration of m_1 is a/2.
For m_2, $T-m_2g=m_2a$ and for m_1, $m_1g-2T=m_1a/2$. Solving the two equations, I get
a=\frac{m_1-2m_2}{2m_2+m_1/2}g
which is wrong.

 

[rl.bhat]

Even though m1 is greater than m2, because of the stretched spring m2 moves downwards. Taking up positive down negative sign, rewrite your two equations and solve for a2.

 

[consciousness]

Sorry for such a late reply, I had some problems with my internet connection.



Ugh..that's the biggest problem I face in these type of problems. If the acceleration of m_2 is a, then acceleration of m_1 is a/2.
For m_2, $T-m_2g=m_2a$ and for m_1, $m_1g-2T=m_1a/2$. Solving the two equations, I get
a=\frac{m_1-2m_2}{2m_2+m_1/2}g
which is wrong.

Using the approach I gave in the 3rd post I am getting the answer

a=g\frac{m_1-2m_2}{2m_2} (upwards)

Can you please tell me if this is the correct answer? I have also used another method given by my teacher to get the same answer.

 

[consciousness]

I think i understood why the constraint relation approach doesn't work. It might have something to do with the fact that objects at rest can also have accelerations(like a vertically thrown ball at its highest point). For an object at rest its acceleration can only be found by Newtons second law.

 

[Saitama]

Using the approach I gave in the 3rd post I am getting the answer

a=g\frac{m_1-2m_2}{2m_2} (upwards)

Can you please tell me if this is the correct answer? I have also used another method given by my teacher to get the same answer.

That's the correct answer.

What's the other method have you used? One of my friend also suggested me the same way as you did in your post #3.

 

[consciousness]

That's the correct answer.

What's the other method have you used? One of my friend also suggested me the same way as you did in your post #3.

The other method is that when the string is burnt the NET force on m2 will be the force it applied on m2 before it was burnt but in opposite direction (ie upwards). Now we can get the answer in one equation very quickly. This result is very useful and can be easily derived also.

Oh and its adivisible to ignore my post #18 i feel that the logic is flawed now.

 

[PhanthomJay]

I think i understood why the constraint relation approach doesn't work. It might have something to do with the fact that objects at rest can also have accelerations(like a vertically thrown ball at its highest point). For an object at rest its acceleration can only be found by Newtons second law.

Yes, per your example of the thrown ball, an object can have an acceleration even when temporarily at rest.

Using the approach I gave in the 3rd post I am getting the answer

a=g\frac{m_1-2m_2}{2m_2} (upwards)

Can you please tell me if this is the correct answer? I have also used another method given by my teacher to get the same answer.

Well, interesting, your solution assumes the same initial tension in the spring and cord before and after the cut,that is ,T = (m_1)(g)/2. When you look at a free body diagram of m_2, and use Newton 2 and set T- m_2(g) = m_2(a_2), its acceleration, a_2, upward, is as you have stated. But when you look at a free body diagram of m_1, then m_1(g) - 2T = m_1(a_1), from which, a_1 = 0! Thus, their initial accelerations are not related by 2a_1 = a_2, which would be the case if there were no spring in the problem and the cords were inextensible.

 

[consciousness]

Yes, per your example of the thrown ball, an object can have an acceleration even when temporarily at rest.

Well, interesting, your solution assumes the same initial tension in the spring and cord before and after the cut,that is ,T = (m_1)(g)/2. When you look at a free body diagram of m_2, and use Newton 2 and set T- m_2(g) = m_2(a_2), its acceleration, a_2, upward, is as you have stated. But when you look at a free body diagram of m_1, then m_1(g) - 2T = m_1(a_1), from which, a_1 = 0! Thus, their initial accelerations are not related by 2a_1 = a_2, which would be the case if there were no spring in the problem and the cords were inextensible.

The constraint relation makes some false assumptions but I am not sure what. Maybe when the spring stretches by a very small distance then the length is almost constant but for instantaneous acceleration the very small distance plays a huge role as it is defined for small distances.

 

[PhanthomJay]

Yeah, different results when you go from ideal inextensible cords to real world elastic cords and springs.