Calculating the Reading on Newton Meter with 3 Kg Masses

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SUMMARY

The discussion centers on calculating the reading of a Newton meter in a pulley system with a 3 kg mass under gravitational acceleration of 10 m/s². The correct reading on the Newton meter, when accounting for the acceleration of the masses, is definitively 27 N, not 30 N as initially suggested by a teacher. The calculations involve applying Newton's second law (F=ma) and considering the tension in the string, leading to the conclusion that the total reaction force on the meter is 2T, where T is the tension calculated to be approximately 13.33 N.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force, mass, and acceleration
  • Ability to analyze pulley systems and tension in strings
  • Knowledge of free body diagrams for visualizing forces
NEXT STEPS
  • Study the principles of Atwood's machine for further insights on pulley systems
  • Learn how to derive tension in a string using multiple masses
  • Explore the effects of friction in pulley systems and how it alters calculations
  • Investigate real-world applications of Newton meters in measuring forces
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of pulley systems and force measurements.

temaire
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Homework Statement


A pulley system is hanging from a Newton meter as shown below. If we assume that gravitational acceleration is 10 m/s^{2}, what will the reading be on the Newton meter as the masses accelerate?

a) 3 Kg
b) 13 N
c) 27 N
d) 30 N

http://img219.imageshack.us/img219/8816/pulleysc7.png​
[/URL]

Homework Equations


F_{net}=ma
F_{g}=mg


The Attempt at a Solution


I think that the answer is d), because if you add the masses of the two objects, which turns out to be 3.0 Kg, and multiply that by the gravitational acceleration, you would get 30 N. Am I right?
 
Last edited by a moderator:
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If they were stationary - yes, but the weights are falling!
 
So how would I get the correct answer?
 
Draw a free body diagram and apply F=ma for each mass.
 
I was playing around with some of the numbers and I got 27 N. However, I might have gone wrong somewhere.

http://img292.imageshack.us/img292/314/pulleyworkdf6.png​
[/URL]
 
Last edited by a moderator:
temaire said:
I think that the answer is d), because if you add the masses of the two objects, which turns out to be 3.0 Kg, and multiply that by the gravitational acceleration, you would get 30 N. Am I right?

No.

I presume it's a frictionless pulley. Let T be the tension in the string. Let 'a' be the accn of the two masses, toward the heavier mass.

For the 2 kg mass, 2*a = 2*g - T.
For the 1 kg mass, 1*a = T - 1*g.

The total reaction on the meter is 2T. Find that.
 
Shooting star said:
No.

I presume it's a frictionless pulley. Let T be the tension in the string. Let 'a' be the accn of the two masses, toward the heavier mass.

For the 2 kg mass, 2*a = 2*g - T.
For the 1 kg mass, 1*a = T - 1*g.

The total reaction on the meter is 2T. Find that.

2*a = 2*g -T
2*3.333... = 2*10 - T
6.666... = 20 - T
T = 13.333...

2T = 2*13.333...
T = 26.666...
T = 27 N

Thanks Shooting Star.
 
holy crap i have a problem on my homework just like that. sweet! thanks. except i need the acceleration. my blocks are 5.0kg (block A) and 2.0kg (block B)...help w/ acceleration please!
 
culexor said:
holy crap i have a problem on my homework just like that. sweet! thanks. except i need the acceleration. my blocks are 5.0kg (block A) and 2.0kg (block B)...help w/ acceleration please!

There are two eqns with a and T. What do you think you should do?
 
  • #10
The problem that I posted at the very beginning of this thread was off my memory. Here is how it was actually worded.

http://img229.imageshack.us/img229/9058/dynamics2nj2.jpg

Is the answer still 27N, because when I asked my teacher, she said that the answer was 30N.
 
Last edited by a moderator:
  • #11
Can anyone help?
 
  • #12
EDIT: I became a bit upset to see that you had written T= 26.6 N in post #7, but I guess you meant 2T. So , the answer of 27 N still stands.

Ask your teacher again. The answer cannot be 30 N.
 
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  • #14
If the masses exert a force of 30N downwards, and the net force is 10 Newtons downwards, shouldn't the reaction force at the pulley be about 20N upwards? What is wrong with my answer?

Ok I udnerstand where I was wrong now! haha =X
 
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  • #15
I agree that the answer is 27 N.

I solved it without looking at anyone's answers in the thread, without even looking at the choices for the answers.

The acceleration of the system is 3.33 m/s^2.

The net force on the 2kg mass is ma=6.67 N. This is equal to mg -T, so T = 20 N - 6.67 N= 13.33 N. The scale pulls up on 2*T = 27 N.

Your teacher is incorrect. Point out that the "30N" violates Newton's 3rd Law as well as the Law of conservation of momentum, as well as the Conservation of Energy principle.
 
  • #16
Ah, but if the masses are held, then released to accelerate, the reading will show 30N at that instant, then decrease to 26.6666 and oscillate about that point (since its a spring system). It takes time for the reading to move from 30N to 26.66N.

*at least if I was a teacher who always wanted to be right, I would say this :)
 
  • #17
K.J.Healey said:
Ah, but if the masses are held, then released to accelerate, the reading will show 30N at that instant, then decrease to 26.6666 and oscillate about that point (since its a spring system). It takes time for the reading to move from 30N to 26.66N.

*at least if I was a teacher who always wanted to be right, I would say this :)

As a teacher, I would not waste time splitting hairs with "gotcha" scenarios. It is a good idea to be reminded of what things really do in the real world, but no one is going to be able to calculate anything if we did take all real considerations into account.

And even with real spring scales, there is just enough damping to produce quick and accurate demonstrations of just this problem. I just did it right now with a standard large-display spring scale. It works quite well. I think the OP should ask for a demo!
 
  • #18
K.J.Healey said:
Ah, but if the masses are held, then released to accelerate, the reading will show 30N at that instant, then decrease to 26.6666 and oscillate about that point (since its a spring system). It takes time for the reading to move from 30N to 26.66N.

*at least if I was a teacher who always wanted to be right, I would say this :)

We are not talking about the Physics any more, nor about the rationale behind the answer given by the OP's teacher, the logic of which is nonsense. I have expressed my real concern in the other thread, viz., the fate of future students made to learn something wrong.
 
  • #19
This is just standard Atwood's machine calculation, but the multiple-guess answers are interesting. It seems obvious that the person who constructed the question knew the answer to be 27 N since that is the only reasonable way to have any discrimination in the choice between B and C. Did she "lift" the question from another prof without bothering to get the answer key?
 
  • #20
I think your teacher may not notice the question.
Since the system is accelerated.
 

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