Solved: Find Tension in Strings of Uniform Stick Suspended from Ceiling

In summary, the conversation discusses a problem from the chapter of rotational motion, involving a uniform stick of length 1 meter and mass 200 gm suspended from the ceiling through two vertical strings of equal length fixed at the ends. A small object of mass 20 gm is placed on the stick at a distance of 70 cm from the left end. The problem is solved using different methods such as center of mass distribution and sum of torques. However, there is some confusion regarding the distribution of the 20 gm mass into two parts and the correct value of the tension in the two strings. The conversation concludes with a recommendation to use the sum of torques method and an explanation of how to use it.
  • #1
hellboydvd
13
0
A uniform stick of length 1 meter and mass 200 gm is suspended from the ceiling through two vertical strings of equal length fixed at the ends. A small object of mass 20 gm is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.



The problem i attempted is basically from the chapter rotational motion.

Attempt:

I had the confusion regarding applying centre of mass to distribute both the masses ( of stick and object) at both the ends of the stick, thereby calculating resultant effective mass at both ends. e.g. as the stick is uniform, i distributed 200gm into 100 gm each and placed it at both the ends of the stick, secondly, i distributed the 20 gm mass into two parts such that its centre of mass is at 70 cm from the left end( as given in the problem), so by centre of mass formula i calculated it to be 6 gm and 14 gm to be placed at left and right ends respectively, so i came up with the answer as -: net force on right side=(1/100)*10+(14/1000)*10 = 1.14 Newtons. and on the left similarly, it comes to 1.06 Newtons.

But the answer is given as 1.12 N(right) and 1.04 N(left)
please help clarify my doubt ( or wrong solution)!
 
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  • #2
hellboydvd said:
I had the confusion regarding applying centre of mass to distribute both the masses ( of stick and object) at both the ends of the stick, thereby calculating resultant effective mass at both ends. e.g. as the stick is uniform, i distributed 200gm into 100 gm each and placed it at both the ends of the stick, secondly, i distributed the 20 gm mass into two parts such that its centre of mass is at 70 cm from the left end( as given in the problem), so by centre of mass formula i calculated it to be 6 gm and 14 gm to be placed at left and right ends respectively, so i came up with the answer as -: net force on right side=(1/100)*10+(14/1000)*10 = 1.14 Newtons. and on the left similarly, it comes to 1.06 Newtons.
An interesting approach. One mistake: Redo the distribution of the 20 gm mass into two. (Check the center of mass of your two pieces.)

I recommend using the sum of torques. A bit easier. (But your method is fine, done right.)
 
  • #3
@Doc Al

i checked it twice, still i found it to be the same. the sum of torques method, ah, i tried r x F taking from the left end ( F=mg = 10*20/1000) and thereby rotating it through some angle, but got confused as the whole stick would also rotate around the axis perpendicular to its left end, i got stuck into angular acceleration etc, so can i get a few steps to your approach, thanks for the help.
 
  • #4
The book uses g = 9.8 whereas you used 10.

You did well. There is also an equivalent way of doing it which may seem simpler : write down the force equation and the torque equation. For the system to be in equilibrium, both force and torque must be zero. Plug in the values and you're done.
 
  • #5
hellboydvd said:
i checked it twice, still i found it to be the same.
Try it again: A 6 gm piece at x = 0 and a 14 gm piece at x = 200 cm. Where's the center of mass?
the sum of torques method, ah, i tried r x F taking from the left end ( F=mg = 10*20/1000) and thereby rotating it through some angle, but got confused as the whole stick would also rotate around the axis perpendicular to its left end, i got stuck into angular acceleration etc, so can i get a few steps to your approach, thanks for the help.
Nothing is accelerating; the stick is in static equilibrium so the net torque about any point is zero. To find the tension in the right thread, find torques about the left end of the stick.
 
  • #6
@ Doc Al

sorry, i think u got something wrong there, the stick is 100 cm( 1 metre) long, not 200 cm.
 
  • #7
hellboydvd said:
@ Doc Al

sorry, i think u got something wrong there, the stick is 100 cm( 1 metre) long, not 200 cm.
D'oh! Sorry about that! :rolleyes:

Looks like xboy found your "mistake".

In any case, try using the sum of torques method. It's easy.
 
  • #8
Confusion 1: will the torque due to weight * 70cm ( from left) = torque 1 and second torque is obtained from 30cm*weight?

I'm having a bit trouble in equating torques, can u write it down for getting me started. Thanks!
 
  • #9
When you take torque, you take it about a point. First you have to find some point which you like (which can be any point, but you may want to choose something that makes the calculation simple.) Then you look how far the lines of forces are from this point - you drop perpendiculars to the lines of forces from this point and you look at their lengths. You use this length to calculate torque. You also work out the directions of the torques. This can be done through the thumb rule, or by looking at the forces and 'seeing' which way they are going to rotate the object.

If the system is in equilibrium you equate the total torque to zero. That's the long and short of the torque method.
 
  • #10
hee hee, I though the title of this thread meant it was about people who get tense (in threads like this)
 
  • #11
:smile: :smile:
 

Related to Solved: Find Tension in Strings of Uniform Stick Suspended from Ceiling

1. What is the purpose of finding the tension in strings of a uniform stick suspended from the ceiling?

The purpose of finding the tension in strings of a uniform stick suspended from the ceiling is to understand the forces acting on the system and ensure that the stick is stable and not at risk of falling.

2. What factors affect the tension in the strings?

The tension in the strings is affected by the weight of the stick, the angle at which the strings are suspended, and the strength and length of the strings.

3. How is the tension in the strings calculated?

The tension in the strings can be calculated using the formula T = mg / cosθ, where T is the tension, m is the mass of the stick, g is the acceleration due to gravity, and θ is the angle at which the strings are suspended.

4. Can the tension in the strings be greater than the weight of the stick?

Yes, the tension in the strings can be greater than the weight of the stick if the angle at which the strings are suspended is less than 90 degrees. In this case, the tension must be strong enough to support the weight of the stick and keep it suspended.

5. Why is it important to find the tension in the strings accurately?

It is important to find the tension in the strings accurately because an incorrect tension can result in an unstable system, potentially causing the stick to fall or other safety hazards. Additionally, knowing the tension can provide valuable information for designing and constructing similar systems in the future.

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