1. Feb 17, 2009

### hellboydvd

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)

2. Feb 17, 2009

### Staff: Mentor

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. Feb 17, 2009

### hellboydvd

@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. Feb 17, 2009

### xboy

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. Feb 17, 2009

### Staff: Mentor

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?
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. Feb 17, 2009

### hellboydvd

@ Doc Al

sorry, i think u got something wrong there, the stick is 100 cm( 1 metre) long, not 200 cm.

7. Feb 17, 2009

### Staff: Mentor

Looks like xboy found your "mistake".

In any case, try using the sum of torques method. It's easy.

8. Feb 17, 2009

### hellboydvd

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. Feb 17, 2009

### xboy

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. Feb 17, 2009

### sirchasm

hee hee, I though the title of this thread meant it was about people who get tense (in threads like this)

11. Feb 17, 2009

### xboy

:rofl: :rofl: