# Thread: Friction on an incline

1. ## Friction on an incline

A loaded penguin sled weighing $80 \ \mbox{N}$ rests on a plane inclined at angle $\theta=20^{\circ}$ to the horizontal. Between the sled and the plane, the coefficient of static friction is $0.25$, and the coefficient of kinetic friction is $0.15$.
(a) What is the least magnitude of the force $\overrightarrow{F}$, parallel to the plane, that will prevent the sled from slippig down the plane?
(b) What is the minimum magnitude $F$ that will start the sled moving up the plane?
(c) What value of $F$ is required to move the sled up the plane at constant velocity?

Thankyou... I'm having a hard time grasping the concept of static & kinetic friction.

2. Nobody? Can someone at least post their answers? Thanks

3. Don't know if it's too late but :

a) If you push the sled up the incline, then the force you're applying + the frictional force opposing the force down the plane must counteract the component of the force of gravity parallel to the incline, i.e.
$F_{g ||} = F_{a} + F_{fr}$

$80\cos20^{o} = F_{a} + \mu_{static}F_{N} \quad \quad \mbox{Be careful about which coefficient you're using.}$

$80\cos20^{o} = F_{a} + (0.25)(80\sin20^{o})$
etc.

b) Friction opposes motion so it must be in the opposite direction as it was in a. Since it is on the verge of moving, static friction has reached its maximum just as it is about to move. So:
$F_{a} = F_{g ||} + F_{fr}$

$F_{a} = 80\cos20^{o} + \mu_{static}F_{n}$

$F_{a} = 80\cos20^{o} + (0.25)(80\sin20^{o})$

c) Since it's moving at constant velocity up the incline, $F_{net} = 0$ (so the magnitudes of the opposing forces will be equal) and you will be using the coefficient of kinetic friction.
$F_{a} = F_{g ||} + F_{fr}$

$F_{a} = 80\cos20^{o} + \mu_{kinetic}F_{n}$

$F_{a} = 80\cos20^{o} + (0.15)(80\sin20^{o})$

Hope I didn't make any mistakes. Kind of rusty on the physics here.