# 3 balls in a moving mechanics problem

• Manolisjam
In summary, the problem involves a system of 2 balls, A and B, connected by a rope to a third ball, G. The mass of G is twice that of A and B. The system is moving due to the effect of G's mass and its projection on the line AB, with no friction. The task is to find the time of impact and the velocity of G at that time, with A and B only moving on the x-axis. The solution involves considering two free body diagrams for G and A, using energy conservation, and deriving the relation between their accelerations through the angle θ. The tensions on the rope are not constant and are equal but not equal to mg due to symmetry.
Manolisjam

## Homework Statement

Consider 2 balls A,B on the same line . and they are connected to a third one G with a rope L. AG, AB. now the system monves in the effect of the mass of G and its projection to the line AB is in the middle. No friction. mass of A=mass of B=m and mass of G=2m
.FInd the time of impact and the velocity of G at that time A-B move only on the x axes.

## The Attempt at a Solution

Dont understand what it means the system moves on the effect of the mass G. Also I know the only forces are From the rope. 2 on the G ball and 1 in each of the balls.I understant i have to apply conservation law of energy but how? What is the potential energy(what if the force) ?

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Manolisjam said:
Dont understand what it means the system moves on the effect of the mass G.
I think it means that mass Γ is pulling the the other two masses towards each other as it drops down. If masses A and B are moving horizontally does their potential energy change? What about mass Γ?

kuruman said:
I think it means that mass Γ is pulling the the other two masses towards each other as it drops down. If masses A and B are moving horizontally does their potential energy change? What about mass Γ?
so what is the force acting on G 2mg?

Manolisjam said:
so what is the force acting on G 2mg?
Yes, plus the tension from each string.

kuruman said:
Yes, plus the tension from each string.
So far so good?

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Manolisjam said:
So far so good?
Nope. Why is T1 = mg = T2? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?

kuruman said:
Nope. Why is T1 = mg = T2? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?
From symmetry? ALso why isn't the tension constant? the rope isn't elastic .

Manolisjam said:
From symmetry? ALso why isn't the tension constant? the rope isn't elastic .
my Tx isn't constant

kuruman said:
Nope. Why is T1 = mg = T2? Furthermore, the tension is not constant as mass G descends. If I were you, I would stick to the original idea of energy conservation and not try to derive the equation of motion. Note that the time of impact depends on how far apart masses A and B are when they are released (I assume) from rest. Does the problem give that distance?
IM really confused...

Manolisjam said:
From symmetry? ALso why isn't the tension constant? the rope isn't elastic .
Symmetry demands that the tensions be equal, T1 = T2, but not that they are equal to mg which is what you had.
Manolisjam said:
my Tx isn't constant
I agree, but neither Ty is constant. Is the acceleration constant or does it depend on θ?

kuruman said:
Symmetry demands that the tensions be equal, T1 = T2, but not that they are equal to mg which is what you had.

I agree, but neither Ty is constant. Is the acceleration constant or does it depend on θ?
neither Ty is constant what does that say? Also how do you break up a force equal=2mg to 2 equal forces? ANd the acceleration depends on θ

Manolisjam said:
neither Ty is constant what does that say? Also how do you break up a force equal=2mg to 2 equal forces? ANd the acceleration depends on θ
You have to consider two free body diagrams, one for the hanging mass G and one for A. You get two equations of motion. Note that the accelerations of A and G are different but related through angle θ. That relation you have to derive starting from y = (L2 - x2)1/2.

kuruman said:
You have to consider two free body diagrams, one for the hanging mass G and one for A. You get two equations of motion. Note that the accelerations of A and G are different but related through angle θ. That relation you have to derive starting from y = (L2 - x2)1/2.
dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.

Manolisjam said:
dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.
A, B only move in x don't fall

Manolisjam said:
A, B only move in x don't fall
why T1=T2=mg is wrong?

Manolisjam said:
dont get it. COuld you please post a pic with a complete answer I am trying 2 days now.
Sorry, it's against forum rules to post complete answers.
Manolisjam said:
why T1=T2=mg is wrong?
Because in the vertical direction, the forces acting on mass G are 2Ty up and 2mg down. What is the net force on G? What is the net force equal to according to Newton's 3rd law?

kuruman said:
Sorry, it's against forum rules to post complete answers.

Because in the vertical direction, the forces acting on mass G are 2Ty up and 2mg down. What is the net force on G? What is the net force equal to according to Newton's 3rd law?

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You don't really need to find T. You want to write two equations involving T the masses, and the accelerations. These would be Fnet = Ma in the vertical and horizontal directions. So in the vertical direction, you get 2T cosθ - 2mg = - 2maG. Can you write an analogous equation for the horizontal direction?

kuruman said:
You don't really need to find T. You want to write two equations involving T the masses, and the accelerations. These would be Fnet = Ma in the vertical and horizontal directions. So in the vertical direction, you get 2T cosθ - 2mg = - 2maG. Can you write an analogous equation for the horizontal direction?
mind that A-B don't move on the y axes

Manolisjam said:
mind that A-B don't move on the y axes
ALso is it possible to find T without knowning a_g ? just asking cause i want to undersand what is happening

Manolisjam said:
ALso is it possible to find T without knowning a_g ? just asking cause i want to undersand what is happening
Also i found Ty on an angle θ' not on θ

Manolisjam said:
mind that A-B don't move on the y axes
I understand that. However, there is horizontal tension Tx that accelerates A and B in the horizontal direction.
No
Manolisjam said:
ALso is it possible to find T without knowning a_g ?
No, because they are related through Newton's 2nd Law equation. What's happening is that as mass G descends, it pulls the other masses together until they collide.

Manolisjam said:
Also i found Ty on an angle θ' not on θ
It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.

kuruman said:
It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.

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kuruman said:
It doesn't matter because they are related, θ + θ' = 90°. You can choose either one.
i osted my progress so far.

Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?

kuruman said:
Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?
so all i did is a waste?

kuruman said:
Let's say the gravitational potential energy U is zero when all three masses are at the same height and mass G is at the origin O. What is the potential energy when the distance between A and B is 2x and mass G is distance y below the origin? How are x and y related?
ive been trying 2 days now straight .l . I've asked other people no1 will give me a full answer so i can study it and understant it and learn from it . because that the policy .. i give up...

Manolisjam said:
ive been trying 2 days now straight .l . I've asked other people no1 will give me a full answer so i can study it and understant it and learn from it . because that the policy .. i give up...
I am sorry you feel that you should give up. In this forum we do not give answers but guide people to the answers so they could learn how to do things on their own. I tried pointing you in the right direction when I posted #26, but instead of following my lead and answering my questions, you became concerned about your "wasted" time. So be it, I will not waste any more of your time.

kuruman said:
I am sorry you feel that you should give up. In this forum we do not give answers but guide people to the answers so they could learn how to do things on their own. I tried pointing you in the right direction when I posted #26, but instead of following my lead and answering my questions, you became concerned about your "wasted" time. So be it, I will not waste any more of your time.
sorry for being rude I am just frustrated.. really appreciate you wasting your time with me.

I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.

kuruman said:
I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
are you sure conservation will lead me to finding the time of impact?

Manolisjam said:
are you sure conservation will lead me to finding the time of impact?
some1 else told me to llok at the relaton between the accaleration o A in x axes withs its component to the direction AG and then the relation between Gs accelaration and that component

kuruman said:
I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
Ok I am trying to do what you said i get the same with i did already .what is different? MY potential energy was zero when G was at L distance so if its zero at the same height i sa the final potential energy is the initial i found.

kuruman said:
I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
Laso could you classify the diffculty of the problem 1-10. for a math undergrad .

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