Solve Friction Problem 2: Mass 53.75kg, Coeff. 0.86

[-Physician]

Homework Statement

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 ?

 

[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.

 

[-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?

 

[Shootertrex]

So we would have to do with constant velocity

If the force applied to accelerate the desk is constant, then the velocity would not be constant.

So We would multiply coefficient with the mass and gravity of 9.81 and that would be the force?

Yes, any force greater than the force calculated will cause an acceleration.

 

[-Physician]

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

 

[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

 

[Shootertrex]

Actually, if $f=F$ which is $F_{net}=0$, then acceleration is 0

I had already stated this in my first post.

and the velocity is constant if we accelerate the body $F>f$

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.

Make sense?