# Friction problem 2

1. Feb 16, 2012

### -Physician

1. The problem statement, all variables and given/known data
A desk has a mass of 53.75 kilograms. If the coefficient of static friction between the desk and the floor is 0.86, what force must be used to move the desk from rest?

2. The attempt at a solution
$V_0=0$
$m=53.75kg$
$u=.86$
$f=uN=umg=.86(53.75kg)(9.8\frac{m}{s^2})=453.005N$
$F$net$=ma$
$F-f=ma$
$F-453.005N=53.75kg χ a$
If I don't have the acceleration, how can I sort this ?

2. Feb 16, 2012

### Shootertrex

If a force that has the same magnitude as the maximum frictional force is applied to the desk, then $F_{net}=0$. This means that any force applied that is greater will accelerate the object, ie move it.

3. Feb 16, 2012

### -Physician

So we would have to do with constant velocity and the force of friction would be equal to the applied force. So We would multiply coefficient with the mass and gravity of 9.81 and that would be the force?

4. Feb 16, 2012

### Shootertrex

If the force applied to accelerate the desk is constant, then the velocity would not be constant.
Yes, any force greater than the force calculated will cause an acceleration.

5. Feb 16, 2012

### -Physician

Actually, if $f=F$ which is $F_{net}=0$, then acceleration is 0 and the velocity is constant if we accelerate the body $F>f$

6. Feb 16, 2012

### richyr33

Frictional force = μ * total reaction force

total reaction force = mass * gravitational force

You have worked out the force already in your attempt at a solution

7. Feb 16, 2012

### Shootertrex

Actually, if $F>f$ then the object will have a constant acceleration. If after the object has begun to move that $F=f$ then the velocity will be constant.