# Torque equilibrium

1. Dec 6, 2005

### heelp

1. Explain how it is possible for a large force to produce a small torque, and how it is possible for a small force to produce a large torque?

2. Find the tension force in the cord supporting the meter stick for each part of the experiment. Explain how the first condition of equilibrium 1 is satisfied for the meter stick in each part of the experiment?

3. What is the condition for the static equilibrium of a rigid body?

What is the thinking process that goes into solving thse type of problems?

Last edited: Dec 6, 2005
2. Dec 6, 2005

### Astronuc

Staff Emeritus
One must understand the definitions and relationships of the variables involved.

For example, a torque, T, is developed by a force F being applied at some moment, or distance, d, from the axis of rotation, or T = F*d.

For a static situation, i.e. no motion, the net forces and net moments must be 0.

3. Dec 6, 2005

### ranger

For number 3, do you know what static equilibrium is? http://www.rwc.uc.edu/koehler/biophys/2h.html [Broken]
EDIT: Astro beat me to it. Damn his beard

Last edited by a moderator: May 2, 2017
4. Dec 6, 2005

### dicerandom

You need to think about what a torque is and how it behaves. Most of what you need to know can be gotten from studying two equations:

$$\vec{\tau} = \vec{r}\times \vec{F}$$

and:

$$\frac{d \vec{L}}{dt} = \vec{\tau}$$

Edit: got beat to the punch! ;)

5. Dec 6, 2005

### heelp

So is the large force able to produce a small torque because of the mass of the object?

6. Dec 6, 2005

### Astronuc

Staff Emeritus
Torque or linear force are completely independent of an objects mass. Rather the angular or linear acceleration depends upon the applied torque or linear force, respectively, by virtue of the mass - Newton's second law.

F = ma, or F/m = a.

The relationship for torque $\tau$ and angular acceleration $\alpha$ are a little more complicated since the distribution of mass is a factor, and the relationship of torque to angular acceleration is given by

$\tau$ = I $\alpha$, where I is the moment of inertia.

I find this a nice summary of relationships - http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

7. Dec 6, 2005

### dicerandom

The mass of the object doesn't come into the torque equation, it's simply a force times a distance. If we were talking about angular acceleration then a large torque would produce a small angular acceleration if the moment of inertia of the object was large.

Edit: Beat a second time!

8. Dec 6, 2005

### Astronuc

Staff Emeritus
On the other hand, you provided the vector form, which is an important concept to understand, and which is difficult for some to grasp the first time around.

And as ranger indicated, one needs to describe more about the string and stick.

Besides - great minds think alike. Thank you for your contributions gentlemen!

9. Dec 6, 2005

### heelp

I understand t = f*d but where is that distance coming from.

10. Dec 6, 2005

### andrewchang

the distance between the force and the axis of rotation?

11. Mar 16, 2011

### Delphi718

With the question............. Find the tension force in the cord supporting the meter stick for each part of the experiment. Explain how the First Condition of Equilibrium (1) is satisfied for the meter stick in each part of the experiment.

Mass of a suspension clip 16.5 g
Do not forget to include the mass of the suspension clip in the magnitude of each hanging mass.

Table 1. Determination of unkown mass

Setup Lever arm r1, cm Mass m1, g Lever arm r2,cm Mass m2, g Unknown mass, g % Difference
munknown from equation (8) Experimental mass
1 35 236.5 14.5 570.9 504.4 500 0.88%
4 30 336.5 17.7 570.3 503.8 500 0.76%

Table 2. Determination of the lever arm

Setup Mass m1, g Lever arm r1, cm Mass m2, g Lever arm r2, cm Mass m3, g Lever arm r3, cm % error
Experimental From equation (11)
1 86.5 40 136.5 20 186.5 33 33 0.57%
4 186.5 40 186.5 20 286.5 39 39 0.15%

Table 3. Determination of the center of gravity

Meterstick Mass of meterstick m, g Mass m1, g Distance between support and the left end, d, cm Lever arm r1, cm Lever arm r2 from equation (13), cm Location of the center of gravity, d+r2
Uniform Meterstick 139.9 186.5 30 15 19.99642602 49.99642602
Non-Uniform meterstick 170 186.5 25 18.5 20.29558824 45.29558824

Average for the location of the center of gravity for the uniform meterstick 50.08
Average for the location of the center of gravity for the non-uniform meterstick 45.3

Where do I start ????