Question

Physics 12 challenge question?


Is magnitude of Ff(static) maximum 1.4 times the Ffk?

Answer 1

I tested this, I just want to make sure.


Angle of kinetic friction = 21 degrees
angle of static friction = 28 degrees

\[\mu = \sin \theta \div \cos \theta = \tan \theta\]

Answer 2

plug in

tan (21) = 0.384
tan (28) = 0.532

\[0.532\div 0.384 = 1.39 \]

Answer 3

So close!

So please tell me does this situation apply to all the situation or it might differ?

I am in physics 12, so explain at my level. :) Thank you

Answer 4

@matt101 if you come online. Help me with this. Thx

Answer 5

If I've understood your question, no, this doesn't apply to all situations. Both the coefficient of static friction and kinetic friction values are always unique to the material being examined, and you can't calculate one given the other. The only thing that can be said is that the coefficient of static friction will always be greater than the coefficient of kinetic friction for a particular material (but not by how much).

Answer 6

I would also add that the equation you have above isn't true for all values of θ. You have to remember that the equation for the coefficient simplifies to tanθ ONLY when there is no net force (i.e. the object is either stationary or sliding at a constant velocity). If you increase the incline, for example, the force of gravity parallel to the ramp will eventually overcome the force of kinetic friction, and you well have an net force causing acceleration down the ramp, such that you have:

\[F_{net}=F_G-F_f\]\[ma=mg \sin \theta - \mu_k mg \cos \theta\]\[u_k=\frac{g \sin \theta -a}{g \cos \theta}\]\[u_k=\tan \theta -\frac{a}{g \cos \theta}\]
You can see now that the expression for calculating the coefficient of kinetic friction is somewhat more complicated because now in addition to tanθ you need term to correct for the additional acceleration (a) coming from the net force.