Static/Kinetic Friction Problem

by Spartan Erik
Tags: friction, static or kinetic
 P: 31 1. The problem statement, all variables and given/known data "A block that weighs 50N is initially at rest on a rough horizontal surface. A 20N horizontal force is then applied to the block. If the coefficients of friction (static = 0.6, kinetic = 0.3), what is the magnitude of the frictional force on the block?" 2. Relevant equations Frictional force = coefficient of kinetic friction x normal force Max static frictional force = coefficient of static friction x normal force 3. The attempt at a solution Once again I'm worried this problem is misleadingly easy: Max static frictional force = coefficient of static friction x normal force Max static frictional force = 0.6 x (50N - 20N) Max static frictional force = 0.6 x 30N Max static frictional force = 18N Is this method correct?
HW Helper
P: 5,346
 Quote by Spartan Erik 1. The problem statement, all variables and given/known data "A block that weighs 50N is initially at rest on a rough horizontal surface. A 20N horizontal force is then applied to the block. If the coefficients of friction (static = 0.6, kinetic = 0.3), what is the magnitude of the frictional force on the block?" 2. Relevant equations Frictional force = coefficient of kinetic friction x normal force Max static frictional force = coefficient of static friction x normal force 3. The attempt at a solution Once again I'm worried this problem is misleadingly easy: Max static frictional force = coefficient of static friction x normal force Max static frictional force = 0.6 x (50N - 20N) Max static frictional force = 0.6 x 30N Max static frictional force = 18N Is this method correct?
No.

You need to draw a force diagram on the block.

What are the normal forces?

What is the horizontal forces.

If the block weighs 50N what is the maximum static frictional force it can resist?
 P: 31 I think I just realized my mistake.. The normal force of the block is the given 50N, and the horizontal force applied is 20N, so Max static frictional force = 0.6 x 50N Max static frictional force = 30N The max static frictional force exceeds the force applied horizontally (30N > 20N)
HW Helper
P: 5,346

Static/Kinetic Friction Problem

 Quote by Spartan Erik I think I just realized my mistake.. The normal force of the block is the given 50N, and the horizontal force applied is 20N, so Max static frictional force = 0.6 x 50N Max static frictional force = 30N The max static frictional force exceeds the force applied horizontally (30N > 20N)
I think you are on to something.

Good Luck.
 P: 31 So if the max static frictional force is 30N, and you are only applying 20N of force to it, won't that make the magnitude of frictional force 20N (since the block won't move anyway)?
HW Helper
P: 5,346
 Quote by Spartan Erik So if the max static frictional force is 30N, and you are only applying 20N of force to it, won't that make the magnitude of frictional force 20N (since the block won't move anyway)?
Yes indeed.
 P: 31 Thanks I appreciate it!

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