# physics equations

could anybody tell me the physics equations for friction and torque because i am a stupid algebra two person designing a totaly new drive system and i want to have the math to show it is better. I do have help on doing them though.

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Torque is equal to force x distance.
Friction is equal to the coefficient of friction (mu, a unitless experimental value) x the normal force.

Torque = Force * distance of lever arm from the point of force to the axis of rotation

Friction = Coefficient of friction (represented by the greek letter mu) * normal force

Depending on whether the object is moving or stationary determines the coefficient you use…Static coefficient for stationary objects, kinetic for moving objects.

*Originally posted by Jim McGeehin *
**Torque = Force * distance of lever arm from the point of force to the axis of rotation **

To be a little bit more specific - Torque is a cross product. So, it is the distance from the axis of rotation to a point perpedicular to the force vector. Sorry to complicate things, but it can be very important in many applications.

Raul

I thank ya’ll for the help with the physics equations my sponsor ( a physics teach) rattled them off at me and it kinda went over my head.

*Originally posted by Argoth *
**I thank ya’ll for the help with the physics equations my sponsor ( a physics teach) rattled them off at me and it kinda went over my head. **

So ask again, and write it down

the distance from the axis of rotation to a point perpedicular to the force vector.

I’m confused

points can be perpendicular?

*Originally posted by michael_obrien *
**I’m confused

points can be perpendicular? **

If one were to draw a line from the axis of rotation to the point the force is acting upon, that line would be perpendicular to the component of the force vector exerting the torque.

Sometimes an example can be worth a thousand words (especially in physics).

Consider a rectangular body (represented by periods), that is acted on by a force F, and free to rotate about the point O.

``````
......---> F
.    .
.    .
.    .
O.....

``````

There are two ways to go about computing torque (which lead to the same thing). Torque = Force * moment arm. So to form the “moment arm”, construct a “line of force” (a line paralell to the force vector), and drop a perpindicular line down to the point of rotation. This “perpindicular” is the moment arm, labeled ‘m’ below. The magnitude of the torque is therefore this value m multiplied by the magnitude of the force (torque = Fm). Note that this is a scalar equation.

``````
---------> F
|    .
m |    .
|    .
O.....

``````

Alternatively, you could construct a vector whose initial point corresponds to O, the point of rotation, and whose final point corresponds to the point of application of the force F. You would then take the cross product of the vector F and the “OF” vector. Note that this is a vector equation.

Frictional force=mu* weight.

This works between surfaces, although beware of which mu value you have. Mu (Kinetic) is the frictional coefficient during motion, and Mu (Static) is for beginning motion.

*Originally posted by Zil709 *
*Frictional force=mu weight.
**

bzzzzz wrong.

Frictional Force is mu * Normal force

The normal force is the force of the surface reacting against the robot, on a level surface this is equivalent to the object weight, but on a ramp or incline, this normal force is different. (The weight is split into some component forces).

This difference is an IMPORTANT distinction to make.

For more information contact your friendly neighborhood Physics Teacher/Professor/TA or pretty much any engineers you might know (although, they might have problems actually explaining it to you ;)).

John
Clarkson University PH131 Teaching Assistant

Torque = movie with Ice cube due out next year. http://torquemovie.warnerbros.com/

(When i saw this preview at the end i said “Way to ruin a perfectly good mechaical word.”)