Static/Kinetic Friction Problem

[Spartan Erik]

Homework Statement

"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?"

Homework Equations

Frictional force = coefficient of kinetic friction x normal force
Max static frictional force = coefficient of static friction x normal force

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?

 

[LowlyPion]

Homework Statement

"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?"

Homework Equations

Frictional force = coefficient of kinetic friction x normal force
Max static frictional force = coefficient of static friction x normal force

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?

 

[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)

 

[LowlyPion]

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.

 

[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)?

 

[LowlyPion]

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.

 

[Spartan Erik]

Thanks I appreciate it!