- #1

Gwozdzilla

- 81

- 0

## Homework Statement

A steel washer is suspended inside an empty shipping crate from a light string attached to th top of the crate. The crate slides down a long ramp that is inclined at an angle of 37° above the horizontal. The crate has mass 180kg. You are sitting inside (with a flashlight); your mass is 55kg. As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate hen the string makes an angle of 68° with the top of the crate. What is the oefficient of kinetic friction between the ramp and the crate?

## Homework Equations

ƩF=ma

f

_{k}=μn

## The Attempt at a Solution

Let m

_{c}= mass of crate

and m

_{y}= mass of you

and m = mass of washer

ƩF

_{crate system}= (m

_{c}+ m

_{y})a = (m

_{c}+ m

_{y})gsin(37) - μ(m

_{c}+ m

_{y})cos(37)

Canceling out the combined mass and solving for μ gives:

μ = [gsin(37) - a]/[gcos(37)]

Apparently...

Let T = tension in string

ƩF

_{y-washer}= ma

_{1}= Tsin(22) -mgsin(37)

and

ƩF

_{x-washer}= ma

_{2}= Tcos(22) -mgcos(37)

And a

_{2}= 0 allowing for these two equations to be rearranged and then divided by each other giving...

tan(22) = [gsin(37) - a]/[gcos(37)]

**My actual question:**

Where are the ƩF

_{washer}equations coming from? I have the attached picture to help, but I still don't understand.