Why Is µ Equal to tan θ in Friction Calculations?

[1010]

Homework Statement

Deriv µs = tan θ
For when a block is put on a wooden plank and the plank is lifted at one side.

Homework Equations

tan = sin/cos
Ff = µFn

The Attempt at a Solution

Ff = µFn
-mgsinθ= µ (-mgcosθ)
µ =(-mgsinθ) /(-mgcosθ)
µ = tan θ

Where i am confused is why Ff = -mgsinθ. Is it because the block has an acceleration of zero when falling down the plank?

 

[cristo]

Where i am confused is why Ff = -mgsinθ. Is it because the block has an acceleration of zero when falling down the plank?

Yes, the block is assumed not to move, and so the frictional force must balance the component of the weight of the block parallel to the plank.

 

[ogb p]

I would like to clarify that the equation µ = tan θ is not derived from the given scenario of a block on a wooden plank being lifted at one side. This equation is derived from the concept of friction and the relationship between the coefficient of friction (µ) and the angle of inclination (θ) of a surface.

In this scenario, the equation Ff = µFn is used to calculate the force of friction (Ff) between the block and the wooden plank. This equation assumes that the block is not accelerating, as you correctly pointed out. The negative sign in Ff = -mgsinθ indicates that the direction of the force of friction is opposite to the direction of motion.

To clarify, the equation µ = tan θ is derived from the fact that the coefficient of friction is equal to the tangent of the angle of inclination at which an object will start to slide on a surface. This can be seen by rearranging the equation Ff = µFn to µ = Ff/Fn and substituting the values of Ff and Fn in terms of the applied force (F) and the normal force (N) as follows:

µ = Ff/Fn = (Fsinθ)/(Fcosθ) = tan θ

Therefore, the equation µ = tan θ is a fundamental relationship in the study of friction and is not specific to the given scenario. It is important to understand the underlying principles and concepts behind equations rather than just memorizing them. I would suggest further exploring the concept of friction and its applications to gain a better understanding.