Pulley,Work,Velocity Problem

1. Apr 6, 2005

VSCCEGR

Problem:

The system shown consisting of a 20-kg collar A & a10-kg weight B is @ rest, when a constant 500-N force is applied to collar A (a) Determine the Velocity of A just before it hits the ground @ C (b) solve part a replacing the weight B w/a 98.1-NA downward force. Neglect friction and the mass of the pulleys.

(a) Can I start by figuring the weight of A vs. the counter weight. then sumF=ma on the collar then use a constant Accel. Eq. to find Vel. Or do I use some sort of Kinetic Energy Eq's. (I hate these) ??

(sum)F=ma
500-N=200-kg*a
a=25m/s^2
V^2=Vo^2+2aX
V^2=0+2(25m/s^2).6m
V=5.43m/s

Is this right or is there more forces acting on A. (I assumed this because A starts at rest.)

(b) what would be the difference in solving this one?

Any Help Would Be Appreciated.

Last edited: Apr 7, 2005
2. Apr 6, 2005

VSCCEGR

Any Help Would Be Good?

3. Apr 7, 2005

primarygun

Can you tell me what supports Block A?
Ceiling? or the pulley?

4. Apr 7, 2005

VSCCEGR

The Tension in the cable supports A.

Look At the thumbnail, it's a representation of the problem.

5. Apr 7, 2005

PBRMEASAP

Energy conservation is the way to go. You know the amount of work done on the system, 500N * 600mm. This must be equal to the change in mechanical energy of the system. Since A and B are attached by a string of constant length, you know their final speeds are the same. You also know that B has moved 600mm upward.

edit: oh wait, the length of the string between A and B isn't really constant, just the total length. I'll think about it some more.

edit#2: okay, sorry. Weight B moves twice as far and twice as fast as A in order to keep the total length of the string constant. I think that's it. Pulleys confuse me.

Last edited: Apr 7, 2005
6. Apr 7, 2005

VSCCEGR

So am i correct i assuming that the only force doing work in part a on A is the 500-N force? Or do i have to compensate for the 10kg weight and add in the resulting force created by A?

What i mean is does the 10kg weight hold the 20kg weight in equilibrium?
Therefor all i have to account for is the 500-n force.

Last edited: Apr 7, 2005
7. Apr 7, 2005

Staff: Mentor

This system has two pieces: The collar (A) attached to its pulley, and the hanging weight B. Apply Newton's 2nd law to each of these objects. Start by indentifying the forces acting on each. (Don't forget the tension in the rope.)

Figure out how the acceleration of A relates to the acceleration of B.

Combine these equations and you can solve for the acceleration of each object.

In this case, you are given the force applied to the rope, which equals the tension in the rope. (In part a you had to solve to find the tension.)

8. Apr 7, 2005

Staff: Mentor

You can also use energy methods to solve this. The applied force does work, which equals the change in total energy of the system. Note that the movements of objects A and B are coupled: what's the relationship between their speeds and heights?

9. Apr 7, 2005

Staff: Mentor

It looks like you already got some help in the other forum! I merged the two threads.
If you wish to treat A separately, then don't forget the ropes which exert a force on A.

If you consider the entire system together (A and B) then the only external forces are the applied force and gravity.

10. Apr 7, 2005

VSCCEGR

Doc Al your confuseing me more than helping me!
Let's stick to one method instead of trying to figure the problem multible ways at the same time!

It's from a energy and momentum chapter so let's stick with a Princible of Work and Energy setup.

now:
W=Fd=(delta)KE
KE=.5mV^2
F=?

wt.A=20kg
wt.B=10kg
d=.6m
I'm just having trouble finding F. In (a) is it 500-N since it is supposed to be in Equi. just before the 500-N is applied or do I have too account for the pulley set up? If so How, i do not get this pulley setup.

11. Apr 7, 2005

Staff: Mentor

Well... you started by calculating the acceleration. Why did you give up on that method?
Nothing wrong with using this method, as long as you include all the forces that do work on the object.

If you choose to isolate body A, then you must include all the forces that do work on A. I see three: The applied force, the weight of A, and the rope pulling up on the pulley attached to A.

If, as I recommended in my last post, you find the work done on the entire system, not just A, then you won't have to find the tension.

12. Apr 7, 2005

VSCCEGR

You know Doc sending people in circles isnt funny. Especially when the are trying to learn something that they DO NOT understand. I agree with the policy of this forum, that spoon feeding isn't the best way to learn, trust me if I have not made attempts at the problems I post then I will Not Post them. But there are time in which one needs more than verbal commands to do somethingor to learn new techniques. To teach someone something you can't just tell them to go do it. You have to give them instruction on the methods applied! I simply wish to figure out how A is affected by B! How do you find the tension. As i said before i do not understand the setup being used.

13. Apr 7, 2005

Staff: Mentor

Why don't you try doing what I suggest? When I thought you were interested in finding the acceleration, I gave you advice on how to do that. (See my first post in this thread.) You start by identifying the forces on both objects. Just do it! Then apply Newton's 2nd law. If there's something that doesn't make sense, just ask.

And, if you would rather use energy methods, both PBRMEASAP and I gave adice about how to do that: $\mbox{Work} = \Delta \mbox{Total Energy}$.

(There is no simple way to uncover the tension in the rope except by solving for it using Newton's 2nd law. At least as far as I can see.)

14. Apr 7, 2005

VSCCEGR

Here is a clue.
I am though, asking for someone to teach me how to do this problem!
WORK=(delta)E
How this problem applies to this however, is.
Now I'll give you this. It's what i understand so far. Is it right? If so then good, i understand. If not how TEACH me something new.

(sum)Fb=ma
T-Wt_B=ma
(sum)Fa=ma
2T-Wt._A=ma

15. Apr 7, 2005

VSCCEGR

Just forget it Doc. I'll go to school tomarrow and get someone who is will ing to TEACH.

I did not want this to get out of controll like this. For heavens sake it is just a physics problem. However when you say that you are going to help someone HELP THEM. DO NOT GIVE THEM THE RUN AROUND.

Last edited: Apr 7, 2005
16. Apr 7, 2005

PBRMEASAP

settle down.

17. Apr 8, 2005

Staff: Mentor

That's OK. I'd write it like this:

$$T - m_B g = m_B a_B$$

You left out the applied force (500N), which I'll call $F_A$:

$$2T - m_A g - F_A = m_A a_A$$

Note that the accelerations of each mass are not equal. The acceleration of B is twice that of A. For simplicity, let's call the acceleration of A "a" downward: so $a_A = -a$. The acceleration of B will be "2a" upward: so $a_B = 2a$. Now plug these into your equations and get:

$$T - m_B g = 2 m_B a$$

$$2T - m_A g - F_A = - m_A a$$

Two equations, two unknowns. Solve for a.