Posted by Bobby C on .
A 32 kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 23.2° the crate begins to slide downward.
(a) What is the coefficient of static friction between the crate and the ramp?
(b) At what angle does the crate begin to slide if its mass is doubled?
I have no idea how to find part a.
For part B why is it not twice the angle?

Physics 
Melissa,
the weight of the box is mg so ( 32x9.81)
In order to find the coefficient of static friction you need the normal force. If you draw out component vectors for the weight you will get:
ycomp: (mg)cos23.2=288.53.
xcomp: mgsin23.2=123.66
If you drew it out correctly you will see that your y component should be equivalent to the normal force ( weight and normal force are equal and opposite, y component ( like normal force) is at a 90 degree angle to the box)
so this means that Normal force= 288.53.
Next I found acceleration using the x component:
123.66=32kg(acceleration)
a=3.86
Now you can set up an equation using Fs as the force
a=Fs/m
3.86=normal(coefficient static)/m
3.86=288.53x/32
x=.427
as for part b I'm still working on it myself...