One for the Weekend (2) - Just for fun - Answer.

Another quiet Friday afternoon, so it's time for another puzzle / problem to occupy those idle hours.

The three masses are attached to the pulleys as shown and are supported underneath initially.

When the support is removed, what will happen to mass C. Will it go up, down or stay where it is?

All other information is in the diagram.

Post your thoughts here if you like.

Answer on Sunday evening.

(edited 13 years ago)

Reply 1

soup

11

On first glance I thought that C would go up as the system on the left had a total mass of 5kg and it would be obvious that C would rise up as 5kg > 4kg.

However I believe that the tensions in the smaller pulley would cancel out and in effect just leave a mass of 1kg on the left and therefore C would fall down at an acceleration of 3/5 g.

Of course I am probably wrong as I am really bad at these kind of puzzles.

Reply 2

boromir9111

17

*Subscribes* shall have a look at this a little later on today!

Reply 3

Xdaamno

Looks like up to me. Those tensions on the left are all internal forces.

Reply 4

Pangol

15

A long and tedious series of equations, which I am sure no-one is interested in unless it comes to a tie-breaker, leads me to think that C will stay where it is.

Reply 5

jaroc

6

Although it is very non-intuitive, I have to agree with Pangol. However, I believe it's possible to show that with just two equations with a good thought experiment :smile:

Reply 6

Xdaamno

I made a bad assumption. A few equations leads me to think C's stationary.

Reply 7

SalBhatti

13

Original post by Xdaamno

I made a bad assumption. A few equations leads me to think C's stationary.

Same here. My equations also lead me to C being stationary. :/

Reply 8

Pangol

15

Spured on my Jaroc's assertion that this only takes two equations, I tried to simplify my argument, and while I've certainly done that, I can't claim to have it as concise as he suggests is possible.

First, just consider A and B passing over the lower pully. Two applications of Newton's Second Law tell us that, if

T_1

is the tension in the string joining A and B and

a_1

is the downwards acceleration of A, then

3g-T_1 = 3a_1

T_1-g = a_1

Adding these together gives us that

a_1 = \frac{g}{2}

, from which we can see that

T_1 = \frac{3}{2} g

.

Now consider the smaller pulley and C. Two applications of Newton's Second Law tell us that, if

T_2

is the tension in the string joining the smaller pulley and C and

a_2

is the upwards acceleration of C, then

g + 3g - T_2 = a_2

T_2 - 4g = a_2

Adding these together gives us that

a_2 = 0

, and so C stays where it is.

Reply 9

Stonebridge

OP

13

Yes. It turns out that C stays where it is. A goes down and B goes up.
Pangol has done the maths in the previous post if anyone is interested.

Reply 10

Azagthoa

12

Can someone explain why it is there only 4g of force acting on the pulley on the left? :smile:

Reply 11

Stonebridge

OP

13

Original post by Azagthoa

Can someone explain why it is there only 4g of force acting on the pulley on the left? :smile:

The reason why it's equal to 4g is really only shown in the maths that you can see in post #9
I take it you mean that the downwards force is 4g and this balances the upwards force of 4g in the tension in the string joining the mass C
This seems to contradict the fact that there is 5kg on the left and 4kg on the right.

Think this way.
If masses A and B were equal and both 1.5kg then the whole system would be in equilibrium and there would be no movement of any masses or pulleys. There would be 4 kg on both sides. Masses A and B together would pull down on the left pulley with a force of 3g. The left pulley is 1g.
If you arrange it so that A and B are not equal, then A and B accelerate up or down due to their different weights. This changes the tension in the string joining them. This change in the tension unbalances the system. It reduces the overall downward force on the left pulley.
Imagine that mass B had no "weight". Mass A would accelerate downwards and there would actually be no tension in the string joining them. The only downwards force on the left would then be the pulley itself.
The larger the difference in weight between masses A and B, the smaller the tension in their string, and the smaller the force pulling down on the left pulley.
It just happens with these mass values that the tension in the string joining masses A and B is such that the two masses pull down on the left pulley with a force of 3g. This combined with the mass of the left pulley equals 4g.
This balances mass C.
But really, the answer lies in the maths and the solution of F=ma applied to the 3 masses and left pulley.

Reply 12

Azagthoa

12

Original post by Stonebridge

The reason why it's equal to 4g is really only shown in the maths that you can see in post #9
I take it you mean that the downwards force is 4g and this balances the upwards force of 4g in the tension in the string joining the mass C
This seems to contradict the fact that there is 5kg on the left and 4kg on the right.

Think this way.
If masses A and B were equal and both 1.5kg then the whole system would be in equilibrium and there would be no movement of any masses or pulleys. There would be 4 kg on both sides. Masses A and B together would pull down on the left pulley with a force of 3g. The left pulley is 1g.
If you arrange it so that A and B are not equal, then A and B accelerate up or down due to their different weights. This changes the tension in the string joining them. This change in the tension unbalances the system. It reduces the overall downward force on the left pulley.
Imagine that mass B had no "weight". Mass A would accelerate downwards and there would actually be no tension in the string joining them. The only downwards force on the left would then be the pulley itself.
The larger the difference in weight between masses A and B, the smaller the tension in their string, and the smaller the force pulling down on the left pulley.
It just happens with these mass values that the tension in the string joining masses A and B is such that the two masses pull down on the left pulley with a force of 3g. This combined with the mass of the left pulley equals 4g.
This balances mass C.
But really, the answer lies in the maths and the solution of F=ma applied to the 3 masses and left pulley.

Thanks for the great explanation!

(edited 13 years ago)

Reply 13

mrCinderella

weight of A=3g
weight of B=1g
3g-1g=2g
weight of small pulley= 2g
total wight of small pulley and A+B= 4g
weight of C=4g
weight of C acts in oposite direction to pulley+A+B therefore resultant=0
done.

Reply 14

Stonebridge

OP

13

Original post by mrCinderella

weight of A=3g
weight of B=1g
3g-1g=2g
weight of small pulley= 2g

The weight of the small pulley is 1g. (Its mass is 1kg)

total wight of small pulley and A+B= 4g
weight of C=4g
weight of C acts in oposite direction to pulley+A+B therefore resultant=0
done.

Reply 15

mrCinderella

Original post by Stonebridge

The weight of the small pulley is 1g. (Its mass is 1kg)

oooops! i fail.

Reply 16

KeyFingot

Original post by Stonebridge

Another quiet Friday afternoon, so it's time for another puzzle / problem to occupy those idle hours.

The three masses are attached to the pulleys as shown and are supported underneath initially.

When the support is removed, what will happen to mass C. Will it go up, down or stay where it is?

All other information is in the diagram.

Post your thoughts here if you like.

Answer on Sunday evening.

Very sorry to resurrect this thread. I've been working through Stonebridge's puzzles and am stuck with this one.

From working on the masses A and B, I get the tension in the string to be 1.5g N and the acceleration of mass A downwards to be 0.5g.

However, when taking into account the forces acting on the 1 kg pulley... If you were to apply Newton's Second Law would the equation you get be something like this. (Call the upwards acceleration of c to be a m/s^2 and the tension in the string towards the pulley T)

3g+g+g-T=a ?

Somebody made a previous post and said the equation would be 3g+g-T=a and I can't see why I am wrong? Help please

Reply 17

bananarama2

Original post by KeyFingot

Very sorry to resurrect this thread. I've been working through Stonebridge's puzzles and am stuck with this one.

From working on the masses A and B, I get the tension in the string to be 1.5g N and the acceleration of mass A downwards to be 0.5g.

However, when taking into account the forces acting on the 1 kg pulley... If you were to apply Newton's Second Law would the equation you get be something like this. (Call the upwards acceleration of c to be a m/s^2 and the tension in the string towards the pulley T)

3g+g+g-T=a ?

Somebody made a previous post and said the equation would be 3g+g-T=a and I can't see why I am wrong? Help please

You've used the masses. What you need to do is consider the fact the forces of the left pulley aren't due to the masses but the tension. So the force on the left rope is the mass of the left pulley plus 2*tension of the lower rope.

Which gives

2*(3/2)g +g -T=a

Reply 18

KeyFingot

Original post by bananarama2

You've used the masses. What you need to do is consider the fact the forces of the left pulley aren't due to the masses but the tension. So the force on the left rope is the mass of the left pulley plus 2*tension of the lower rope.

Which gives

2*(3/2)g +g -T=a

Oh I see. Of course, I was being very silly. Thank you!!