PDA

View Full Version : physics help needed DESPERATELY!


nippongo2001
Dec4-04, 11:22 AM
When you're given the coefficients of both kinetic and static friction, what can you determine from this information, and how do you go about finding it?

For example, I was given a problem in physics that says the following:

A loaded sled weighing 70N rests on a plane inclined at 20 degrees to the horizontal. Between the sled and the plane, the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.11.
a. What's the minimum magnitude of force that will start the sled moving up the plane?
b. What's the value of force required to move the block up the plane at constant velocity?

If there's any way that you can help with any aspect of this problem, I would be permanently in your debt! Thank you!

arildno
Dec4-04, 11:27 AM
Remember:
STATIC friction has the magnitude it needs to have in order to keep the object at rest, up to a MAXIMAL value, given in terms of the static friction coefficient multiplied with the normal force.

Hence, the MINIMUM force required to get the sled moving, is to apply a force equal to the maximum value static friction can have.

For your b) question, the applied force must exactly balance the kinetic force of friction in order to keep the sled's acceleration zero (i.e, its velocity constant).

Doc Al
Dec4-04, 11:31 AM
a. What's the minimum magnitude of force that will start the sled moving up the plane?
To start something moving, you need to overcome static friction.
b. What's the value of force required to move the block up the plane at constant velocity?
Once the block is moving, you have to consider kinetic friction.

nippongo2001
Dec5-04, 10:15 AM
Thank you very much! Your help is indeed appreciated!

However, I am still having difficulty with the actual mathematical process in part A. Is the net force zero? If not, I still come up with two variables at the end. For part A:

In the x direction:
Force_net = F_applied - F_friction - F_parallel
F_net = F_a - (70cos20 x .26) - 70Sin20
F_net = F_a - 17.1028 - 23.9414

Thank you!

Doc Al
Dec5-04, 10:59 AM
However, I am still having difficulty with the actual mathematical process in part A. Is the net force zero?
Yes.
In the x direction:
Force_net = F_applied - F_friction - F_parallel
F_applied = F_friction + F_parallel