Help Needed: Physics Homework Problem with Angle Theta & Acceleration

[piper210_355]

Homework Statement

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.

Homework 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))

The Attempt at a Solution

tan(theta)=.45/F(normal)
i don't know how to find the normal force.

 

[tiny-tim]

Welcome to PF!

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.
…
i don't know how to find the normal force.

Hi piper210_355! Welcome to PF!

Hint: the acceleration perpendicular to the board is obviously zero.

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

No … apply Newton's second law to all the forces along the board.

 

[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?

 

[tiny-tim]

Hi piper210_355!

(have a theta: θ )

tan(theta)=u

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

is that right or am i off?

No, that's fine!

Carry on …

 

[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

 

[tiny-tim]

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

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!