Question

Physics 12 Question!

Answer 1

\[\mu=\frac{ sine \theta }{ \cos \theta } = \tan \theta \]

How is this situation possible?

Answer 2

Are you asking about the identity? Consider the right triangle formed when you rotate a ray through an angle θ around some point. sinθ = O/H and cosθ = A/H. If you then do sinθ / cosθ, you get (O/H) / (A/H) = O/A = tanθ.

Answer 3

do you mean the coefficient of static friction because this is one way of working it out....using an inclined plane

Answer 4

Yes, @Irishboy123 and mat101, it is about inclined planes. My teacher said tan feda = mu

Can you explain why?

Answer 5

Gotcha. Think about a situation where you have a mass on an inclined plane. It can be stationary (in which case μ would be the coefficient of STATIC friction) or it can be sliding downwards along the inclined plane (in which case μ would be the coefficient of KINETIC friction). Whether it's static or kinetic doesn't matter here, as long as we're talking about a case where the net force is 0. When the net force is 0, the force of gravity on the mass acting down the ramp is exactly balanced by the force of friction on that mass acting up the ramp:

\[F_G=f\]
You know that on an inclined plane, the force of gravity on the object PARALLEL to the plane is equal to mgsinθ. You also probably know that friction is μN, where N is the normal force, in this case mgcosθ (the force of gravity on the object PERPENDICULAR to the inclined plane). So in other words, what we have is this:

\[mg \sin \theta= \mu mg \cos \theta\]\[\sin \theta = \mu \cos \theta\]\[\mu= \frac{\sin \theta} {\cos \theta}=\tan \theta\]
Does that make sense?

Answer 6

Yes thank you.