# Homework Help: PHYSICS HELP NEEDeD

1. Oct 12, 2008

### piper210_355

1. The problem statement, all variables and given/known data

Okay so this is probably a really easy problem, but my teacher literally doesn't teach us any thing, so can someone please help me.

The coefficient of static friction between a 2.5kg box and a wooden board is .45 and the coefficient of kinetic friction between the box and the board is .28. The box is placed on the board, which is slowly lifted at one end until the box starts to slide down the board, at which point the angle theta of the board is held constant. Determine the angle theta of the board and the acceleration of the box as it slides down the board.

2. Relevant equations
ok for finding theta i know I can use tan(theta)=F(x)/F(normal)

for acceleration i can use a=(mass)(gravity)(sin(theta))

3. The attempt at a solution
tan(theta)=.45/F(normal)
i don't know how to find the normal force.

2. Oct 12, 2008

### tiny-tim

Welcome to PF!

Hi piper210_355! Welcome to PF!

Hint: the acceleration perpendicular to the board is obviously zero.
No … apply Newton's second law to all the forces along the board.

3. Oct 12, 2008

### piper210_355

ok so i used these equations

F(f) = u*F(n)

F(f) = F(x)

tan(theta) = F(x)/F(n)

and i rearranged them like so to find theta

tan(theta)=F(x)/F(n)
tan(theta)=F(f)/[F(f)/u)]

Here is where I'm not sure if i'm over simplifying(since in the other one i am dividing by a fraction i then change it by multipling by the reciprical)
tan(theta)=F(f)*[u/F(f)]

This cancels out the F(f) leaving...
tan(theta)=u

so then
(theta)=tan^(-1) .45
(theta)=24.23

is that right or am i off?

4. Oct 12, 2008

### tiny-tim

Hi piper210_355!

(have a theta: θ )
No, that's fine!

Carry on …

5. Oct 12, 2008

### piper210_355

Ok I got the rest of these steps from a video on youtube, but I just want to make sure it's right.

First it said to get the force of gravity:
F(g)=(gravity)(mass)
F(g)=(9.81)(2.5)
F(g)=24.5N

Then I found the Force of the parallel (that's what they called the line in between F(n) and F(g) in the video).
F(p)= F(g)sin(24.23)
F(p)=(24.5)(sin 24.23)
F(p)= 10.05

After that I used newton's second law to find the acceleration.
a=F/m
a=(10.05)/2.5
a=4.02m/s^2

6. Oct 13, 2008

### tiny-tim

Hi piper210_355!

Using the web is not a good idea.

Use your books, or lecture notes, and work everything out from basic principles, or you won't be able to do it in the exam.

4.02 would be fine if there was no friction.

Use Newton's second law in the "parallel" direction, and try again!