# Dynamics Problem

1. Oct 3, 2006

### Stevedye56

A crate full of tosters slides down a plam to a loading dock in a warehouse. At the top of the plank the box is given a push so its initial speed is .25m/s. By the time it hits the bottom of the plank, it has a speed of 1.5m/s. The plank is 5.5m long and is set at an angle of 28 degrees. Find the coefficient of friction between the plank and the crate...here's the order of things you need to find out

a) Find the rate of acceleration of the crate.
b) Find the net force on the crate.
c) Find the forece of friction on the crate.
d) Find the coefficient of friction between the plank and the crate.

I was able to calculate the acceleration and got .11 m/s squared.

V_f=1.5m/s
V_o=.25
a=?
d=5.5m

a=V_f^2-V_o^2/2d

Do i use mgcos(theta) and mgsin(theta) to get the perpindicular and paralel components of weight?

If i could figure out how to get F_n i can solve the rest im just at a block which i think is farely simple im just missing some simple thing. Any help would be excellant.

-Steve

EDIT:
I did draw a free body diagram with all the forces: F_N, F_f, w, perpindicular component of weight, and the paralel component of weight

Last edited: Oct 3, 2006
2. Oct 3, 2006

### Saketh

On an inclined plane, $$mg\cos{\theta}$$ is the perpendicular component of weight, and $$mg\sin{\theta}$$ is the parallel component of weight. If you are confused by this, I recommend that you go through the geometry of it until you understand.

When you write your perpendicular net force equation, you will have something like this:
$$\vec{F}_{net perpendicular} = m\vec{g}\cos{\theta} - \vec{F}_N$$
Because the block does not accelerate in the perpendicular direction (i.e. it is not falling off of the plane), we can set the left-hand side of the equation to zero, and find an expression for normal force.
$$\vec{F}_{net perpendicular} = 0 = m\vec{g}\cos{\theta} - \vec{F}_N$$
$$m\vec{g}\cos{\theta} = \vec{F}_N$$

3. Oct 3, 2006

### Stevedye56

that makes perfect sense thanks so much

4. Oct 3, 2006

### Stevedye56

I calculated the F_N=8.82N

10cos28

then to get the frictional force i took

tan28(8.82) which gave me a F_f=4.9N

Is this all correct?

5. Oct 3, 2006

### Saketh

I'm assuming that you derived $$\mu = \tan{\theta}$$ from a zero-acceleration net force equation. However, there is acceleration in this particular case, so you cannot solve for $$\mu$$ by just setting acceleration equal to zero.

$$ma = \mu m g \cos{\theta} - mg\sin{\theta}$$
You are given enough information in the problem to solve for the coefficient of friction. Remember kinematics.

6. Oct 3, 2006

### Stevedye56

I understand that equation. Although i do not understand how to calculate mass. i know that m=w/g but weight is not given, there must be another way...

Last edited: Oct 3, 2006
7. Oct 4, 2006

### Stevedye56

There was no mass given so the problem was either impossible or way too complicated for honors physics