# Relating terminal velocity, force of friction.

1. Jul 12, 2012

### bravoghost

I was working on problems for the MCAT and came across this below. After a few minutes working, I had couldn't come up with the solution. Usually the answer clears up any questions I have, but I'm having trouble figuring out how the force due to gravity matches the force of friction. I thought that the force of friction was something to subtract from the total force available for an object. I don't understand how the force due to gravity equals the force due to friction. How do those two relate? (I understand the mathematics behind it, I just don't understand the reasoning, so please don't just show a bunch of equations and conclude "Here: the math shows it."

Question:
A block slides down a surface at an angle of 30° to the horizontal at its terminal velocity of 5 m/s. If the block masses 12 kg, what is the coefficient of kinetic friction of the surface?

Explanation: As the block travels down the incline at its terminal velocity, the force due to gravity matches the force of friction (hence terminal velocity). The equation, therefore, can be set up as μN = mg(sin θ) where N is the normal force and equal to mg(cos θ). Thus our equation is μmg(cos θ) = mg(sin θ). mg cancels on both sides leaving us with (sin θ)/(cos θ) = μ. Given the angle of 30°, we can calculate that μ = 1/√3 making (B) the correct answer.

2. Jul 12, 2012

### Simon Bridge

When something is moving at terminal velocity, then it is moving at a constant speed, which means that the forces balance so the friction has to balance gravity does't it? Either that or there's another force.

In situations such as air resistance, the friction depends on speed ... so it starts out too small to prevent acceleration so $mg-\mu(v)N=ma$ and increases until it is the same as gravity ... in which case you write $mg-\mu(v_t) N = 0 \Rightarrow mg=\mu(v_t)N$.

Last edited: Jul 12, 2012
3. Jul 12, 2012

### azizlwl

Fnet=ma
a=0 for teminal velocity.

Fg+(-Ffr)=0

Ffr is negative since it opposes the direction of the motion.

4. Jul 12, 2012