Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Saitenji****Member**- Registered: 2005-09-24
- Posts: 10

A crate starts from rest and slides 8.35 m down a ramp. When it reaches the bottom it is traveling at a speed of 5.25 m/s. If the ramp makes an angle of 20.0° with the horizontal, what is the coefficient of friction between the crate and the ramp?

Does anyone know how to solve this? I've tried all that I can to my knowledge, but I've yet to come up with a solution. Of course, I'm guessing that's natural because we're barely starting this unit, but I still want to know how this problem can be solved. If possible, please help me with what steps I should follow...I believe I know the equations, but where to apply them is what I don't. Thanks in advance.

*Last edited by Saitenji (2005-10-05 12:55:39)*

Offline

**kybasche****Member**- Registered: 2005-09-22
- Posts: 8

I'll have a go... someone correct me if I make a mistake.

basically we're looking at something like this...

You'll notice in the diagram that the angle between the gravitational force acting straight down and the normal force on the block (force due to gravity acting perpendicular to the surface of the ramp) is equal to theta, where theta is the angle of the block with the horizontal.

Another picture to help explain that (if you didn't know already)

So... using these relations (note the first picture) and summing all the forces, we can say...

-mu*M*g*cos(theta)+M*g*sin(theta)=a

where 'mu' is the coefficient of friction, 'M' is the mass, 'g' is the gravitational constant, and 'theta' is the angle between the ramp and horizontal

the M cancels out of the equation to leave

g*sin(theta) - mu*g*cos(theta) = M*a [eqn 1]

Looking at the constant acceleration formulas, the best one to use would be

V^2=V_0^2+2*a*(x-x_0) [eqn 2]

where 'V' is the final velocity, 'V_0' is the initial velocity, 'a' is acceleration, 'x' is final location, 'x_0' is initial location

You can see that both 'V_0' and x_0 are zero (starts from rest, and we'll call the starting location zero. You also know 'V' to be 5.25 m/s and 'x' to be 8.35 m from the problem statement. You've solved for 'a' using [eqn 1] and then you can plug 'a' into [eqn 2] to obtain a value for 'mu'.

That's about it... like I said, if I messed something up, someone jump in and save me!

~Derek

Offline

**Saitenji****Member**- Registered: 2005-09-24
- Posts: 10

Thanks so much Derek. I pretty much understood the rest of it, but I missed the part where M cancels out...In fact, no one in my class got the question right.

In any case, this really helps me and I'll keep it in mind for future reference. Thanks again.

Offline

Pages: **1**