# Friction Forces help

1. A 25.0-kg block is initially at rest on a horizontal surface.
A horizontal force of 75.0 N is required to set the block in
motion, after which a horizontal force of 60.0 N is required
to keep the block moving with constant speed. Find (a) the
coefficient of static friction and (b) the coefficient of
kinetic friction between the block and the surface.

2. Homework Equations

3. i need help setting up the problem. i would apply the net force model for the x direction and equilibrium model for the y model. i need to know if this is right.
Fx = 75.0N - F(static) - 60.0N = ma
Fy = n - mg = 0

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rl.bhat
Homework Helper
What is the expression for the coefficient of static friction?

is the expression for static friction just
-Fs = ma

rl.bhat
Homework Helper
is the expression for static friction just
-Fs = ma
Not the above expression.
The expression for μ in terms of normal force and maximum frictional force.

Not the above expression.
The expression for μ in terms of normal force and maximum frictional force.

oh okay. so i got

μ(s)n = ma

than i got n = mg for the second equation.

do i plug in "n" to the expression μ(s)n = ma ?

which would give me μ(s)mg = ma

rl.bhat
Homework Helper
oh okay. so i got

μ(s)n = ma

than i got n = mg for the second equation.

do i plug in "n" to the expression μ(s)n = ma ?

which would give me μ(s)mg = ma
Correct. Now in the given problem, what is the normal force and ma?

Correct. Now in the given problem, what is the normal force and ma?
hmm would mg = -245?
i don't know what the normal force is. i thought that mg and n always canceled to make it equal to zero.

opps. i meant would ma = 75

rl.bhat
Homework Helper
Normal reaction is equal and opposite to mg. mg acts on the floor and normal reaction acts on the block. Now you can find the coefficient of static friction.

Normal reaction is opposite to mg. Now you can find the coefficient of static friction.
THANKS! i found it! 