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Old 10-07-08, 07:34 PM   #1
cuevélo
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Physics- Tension

I have a question.

There's a 5.00kg mass on a horizontal, frictionless table, attached by a light string to a 10.00kg mass. The 10kg mass is off of the side of the table, and there is a light pulley that it is suspended by.

I figured that the 10kg mass is going to have a force of 10*9.8=98N downwards, so the acceleration of both the masses, in their obvious directions will be F=ma=98=15*a --> a=6.53m/s^2.

My question is about the tension in the string between the masses. I think that it will be F=ma-->T=ma--> T=5.00kg*6.53m/s^2--> T=32.65N. Am I right about finding the tension?

P.s. I did draw a free body diagram.
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Old 10-07-08, 07:39 PM   #2
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Here's a picture.
Attached Images
File Type: jpg diagram.JPG (9.2 KB, 0 views)
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Old 10-07-08, 08:16 PM   #3
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I'm not wading through your numbers, and I'd encourage you to derive a solution symbolically anyway, but your general approach seems valid, if somewhat laborious. But question for you... what's the point of your force diagram if forces aren't labeled on it? In particular, if you're solving for tension, shouldn't tension be one of the things labeled? If tension forces were clearly labeled, then it would have been very easy for you to tell whether what you're doing is right. You also wouldn't have to reason through the problem, but rather, could immediately arrive at an algebraic solution by simply balancing forces.
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Old 10-07-08, 08:18 PM   #4
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PS... care to draw a better force diagram, so we can work through the problem the way it was meant to be solved? For now, simply label the masses as m1 and m2 (I don't care which is which), and use g as the acceleration due to the force of gravity.
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Old 10-07-08, 08:38 PM   #5
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That wasn't supposed to be a force diagram, just a picture to show the set up.
I wish that I could go through it with you, but I don't have time- I have to do a lot of English homework.
Thank you for helping me though.
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Old 10-07-08, 08:45 PM   #6
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Are you assuming the pulley is frictionless or are you going to work that derivative into your equation.
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Old 10-07-08, 10:29 PM   #7
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Are you assuming the pulley is frictionless or are you going to work that derivative into your equation.
One thing that has to be learned when taking classes is what to ignore when working a homework problem. Normally, if things like mass and friction of pulleys and cords aren't given in the problem statement, they are to be ignored as being beyond the scope of the problem and of the lesson being taught.
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Old 10-13-08, 09:41 AM   #8
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And then someone introduces you to calculus and the very first problems you work are these old physics nuggets with pulley friction, rope elasticity, and an environmental derivative.
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Old 10-13-08, 09:50 AM   #9
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If it's frictionless, why doesn't the 10 kg object pull the 5kg object off the table?
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Old 10-13-08, 10:36 AM   #10
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It will. But in the meantime, there will be an acceleration to get the objects moving in the correct directions, and a tension on the string. Once both objects are in freefall, or the larger object hits the ground, the tension will be zero. This problem isn't about the end result (both objects on floor), but about the instantaneous state of affairs.
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Old 10-13-08, 10:48 AM   #11
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It will. But in the meantime, there will be an acceleration to get the objects moving in the correct directions, and a tension on the string. Once both objects are in freefall, or the larger object hits the ground, the tension will be zero. This problem isn't about the end result (both objects on floor), but about the instantaneous state of affairs.
I which case, wouldn't that be the same as just hanging a 10kg mass off the end of a string?
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Old 10-13-08, 10:55 AM   #12
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There is an old problem in the Sear's physics books [This textbook was famous for things like Superman jumps off the Empire State building 5 seconds after a child falls over the side, Superman can accelerate at xxxfps, where does he catch the child? The answer was eleven feet below the sidewalk.] A monkey and a bunch of bananas are suspended from each side of a rope threaded through a frictionless pulley suspended from the ceiling. They weigh the same. The monkey tries to get the bananas, describe what happens as the monkey attempts to get to the bananas? It included a diagram and the solution was the monkey just reaches across and grabs a banana....
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Old 10-13-08, 02:55 PM   #13
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I which case, wouldn't that be the same as just hanging a 10kg mass off the end of a string?
No, it's not. If the string's not attached to anything, there's no tension, and the 10 kg mass falls with acceleration -g (if up is defined as positive). If the string's attached to something much heavier ("infinite" weight, like a wall), then the 10 kg mass doesn't go anywhere, and the tension is m*g (where m=10 kg) since it counters the entire gravitational force.

Since the second mass, at 5 kg, cannot be approximated as zero mass or infinite mass (since 5 kg is of the same magnitude as 10 kg), neither extreme approximation works. The 10 kg mass falls more slowly than a free falling body since it is dragging along another mass (but it falls nonetheless), and the reason for this is the tension on the string that attaches the two masses. Intuitively, (10 kg * g) force is being applied on 15 kg of stuff, so you'll get a 2/3g acceleration rate. The remaining 1/3 g is the tension in the string, transferring some of this force from the falling mass to the sliding mass. And this intuitive (but hard to generalize to more complex problems) approach is the one the OP took. (Not the most straightforward way to solve the problem, but technically correct.)
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Old 10-13-08, 03:21 PM   #14
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I'm a physics grad student, so here you go. The main problem is the 10 Kg mass will NOT fall at 9.8 m/ss. Neither is the force on the 5 Kg mass mg for the 10 kg mass. The main point is The masses are attached by a string, so their accelerations are equal.

Call the 10 Gm Mass M1. WHat are the forces acting on M1?

F=M1a
M1g -T = M1a

Where T is the tension in the string. Notice I called DOWNWARD the positive direction, because this is the direction of the accleration.

Now the forces on M2
The only force that matters is the tension
T=M2a

Now you have two equations and two unknowns. In each equation, Tension and acceleration are the same.
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Old 10-13-08, 05:17 PM   #15
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Exactly.
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Old 10-13-08, 08:53 PM   #16
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Since you guys are so graciously answering physics problems, if you have nothing better to do this Monday night, I have a doozie of my own:

Can anyone explain to me how to differentiate and integrate in regards to polar coordinates? My Engineering Dynamics book is awful (I'm sure there is a special place in hell for R.C. Hibbeler), and it's been a long time since I've taken calculus.

Here is the pattern

r = rur

v = vrur + vθuθ

a = arur + aθuθ

where
vr = dr/dt
vθ = r(dθ/dt)

ar = d^2r/dt^2 - r(dθ/dt)^2
aθ = r(d^2θ/dt^2) + 2(dr/dt)(dθ/dt)

so we really get screwed up equations like

(d^2r/dt^2 - r(dθ/dt)^2)ur + (r(d^2θ/dt^2) + 2(dr/dt)(dθ/dt))uθ

I know that part of the problem is the chain rule, but the book seems to use some sort of trick where
dur/dt = (dθ/dt)uθ, but frankly, I don't understand it.

HALP?
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Old 10-20-08, 10:33 AM   #17
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The solution is exceedingly simple; find a intern doing his grad work in physics/engineering, delegate task to them, sit back planning your awards ceremony.
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Old 10-20-08, 10:52 AM   #18
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Here's a picture.
I agree with jschen. Draw a force-body diagram. That will make the problem very simple.

Good luck!
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Old 10-20-08, 10:56 AM   #19
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Since you guys are so graciously answering physics problems, if you have nothing better to do this Monday night, I have a doozie of my own:

Can anyone explain to me how to differentiate and integrate in regards to polar coordinates? My Engineering Dynamics book is awful (I'm sure there is a special place in hell for R.C. Hibbeler), and it's been a long time since I've taken calculus.
Integration by polar coordinates (surface integration) is a double integration on the area or surface. Similarly, its differentiation is a double diff on axes rho and theta (I think).

In either case, I think you convert to Cartesian coordinates and solve that way. Google can help you I've forgotten the rules; it's been three years since I took that course...

Good luck!
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Old 10-20-08, 01:07 PM   #20
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Thanks, I finally figured it out last night. Good thing too, the midterm is tomorrow. The reason for the back and forth between ur and uθ is that ur=cosθi+sinθj in cartesian whereas uθ=-sinθi+cosθj (perpendicular to ur).

d/dt(cosθi+sinθj) = -sinθi+cosθj

and

d/dt(-sinθi+cosθj) = -(cosθi+sinθj)

so dur/dt = uθ

and duθ/dt = -ur
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