Calculating String Tensions in a Hanging Meter Stick

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In summary, a meter stick (L=1m) with mass M=0.409kg is supported by two strings, one at the 0cm mark and the other at the 90cm mark. The tension in the string at 0cm is equal to the total weight of the meter stick (0.409g) and the tension in the string at 90cm is equal to the weight of the meter stick (0.409g) minus the tension. When the string at 0cm is cut, the stick pivots about the 90cm mark and the tension in the remaining string is found by taking torques about that point and considering the forces acting on the stick. This results in an equation for the
  • #1
cuz937100
A meter stick (L = 1 m) with mass M = 0.409 kg is supported by two strings, one at the 0 cm mark and the other at the 90 cm mark.

a) What is the tension in the string at 0 cm?
b) What is the tension in the string at 90 cm?
c) Now the string at 0 cm is cut. What is the tension in the remaining string immediately thereafter?

I got parts a and b figured out, but I can't get c...Can I please get some help...The attachment is a picture of what the problem looks like.
 

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  • #2
Part c is the easiest. The meterstick just becomes a generic weight of 0.409 kg hanging from a string. The tension in the string is just 0.409 g.

- Warren
 
  • #3
No, that's answers not correct...I thought that it would work out like that too, and the tension would be equal to the total weight of the meter stick, but this answer is not right...
 
  • #4
This is what the help for the problem says, but I still don't understand it...

HELP: Draw a FBD to find all the forces and torques. A convenient axis for computing the torques is the position of the remaining string.
HELP: Unlike the previous parts, neither the net force nor the net torque vanishes. The net force leads to an equation for the acceleration of the CM. The net torque leads to an equation for the angular acceleration. The acceleration and angular acceleration are related to each other (HOW?). Put all this together to find the tension.
 
  • #5
Sorry, you're right -- I missed the word "immediately" in the question. I'm afraid I don't have time to figure this one out right now. :sad:

- Warre
 
  • #6
10*0.409g/9 maybe? (i.e 4.454N for g = 9.8)

Ahh, I made the same mistake. :redface:
 
  • #7
no problem, can someone just help me out when you get the chance, i can't get this to work out in any way...I don't know where I'm going wrong...
 
  • #8
By the way, chroot, I think we should be careful, when responding to people's homework questions, to distinguish between "mass" and "weight". I would certainly never say a "weight of 0.409 kg " or refer to tension in terms of grams.
 
  • #9
HallsofIvy said:
I would certainly never say a "weight of 0.409 kg " or refer to tension in terms of grams.
Actually I think chroot was saying the tension would be 0.409kg times g, i.e mg.
 
  • #10
cuz937100 said:
I got parts a and b figured out, but I can't get c...Can I please get some help...The attachment is a picture of what the problem looks like.
When the left string is cut, the stick pivots about the 90cm mark. Take torques about that point:
(1) mgx = I α (Where "x" is the distance from the pivot to the center of mass of the stick; I is the rotational inertia of the stick about the pivot.)
Note that the acceleration of the c.m. is related to α by a = αx. Thus you can solve for "a".

Now consider the forces acting on the stick:
(2) mg - T = ma
This will give you T, the tension in the remaining string.
 
  • #11
I seem to be in a minortiy here, but it appears obvious to me that immediately after cutting string1, string2 is unaffected. So answer in c. is same as answer in b.
 
  • #12
krab said:
I seem to be in a minortiy here, but it appears obvious to me that immediately after cutting string1, string2 is unaffected. So answer in c. is same as answer in b.
Interesting point. I guess it depends on what is meant by immediately. It will take some time for the tension in the string to readjust to the new demand on it.
 
  • #13
HallsofIvy said:
By the way, chroot, I think we should be careful, when responding to people's homework questions, to distinguish between "mass" and "weight."
What I meant was "a generic weight," i.e. a piece of lead, of mass 0.409 kg.

- Warren
 
  • #14
Doc Al said:
When the left string is cut, the stick pivots about the 90cm mark. Take torques about that point:
(1) mgx = I α (Where "x" is the distance from the pivot to the center of mass of the stick; I is the rotational inertia of the stick about the pivot.)
Note that the acceleration of the c.m. is related to α by a = αx. Thus you can solve for "a".

Now consider the forces acting on the stick:
(2) mg - T = ma
This will give you T, the tension in the remaining string.


Thanks for the help everyone, I figured out the answer to the problem...
 

1. How do I determine the weight of the meter stick?

In order to calculate the weight of the meter stick, you will need to know its mass. You can use a scale to measure the mass directly, or you can calculate it using the density and dimensions of the meter stick.

2. What is the formula for calculating string tension in a hanging meter stick?

The formula for calculating string tension in a hanging meter stick is T = mg + ma, where T is the tension, m is the mass of the meter stick, g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration of the meter stick.

3. How do I measure the acceleration of the meter stick?

You can measure the acceleration of the meter stick by using a stopwatch and measuring the time it takes for the meter stick to fall a certain distance. You can then use the formula a = 2d/t^2, where d is the distance and t is the time, to calculate the acceleration.

4. Can I use this method for calculating string tensions for objects other than a meter stick?

Yes, you can use this method for calculating string tensions for any object as long as you know its mass and acceleration. However, the dimensions and density of the object may affect the accuracy of the calculation.

5. Does the angle of the string affect the tension in a hanging meter stick?

Yes, the angle of the string can affect the tension in a hanging meter stick. The greater the angle, the greater the tension will be. This is because the vertical component of the tension force increases as the angle increases.

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