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Beamsbox
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***EDIT***Sorry for a second there, I tried to convince myself that FN was not equal to Fg, but it must.
Given the coefficient of static friction (.4) between an egg and a pan, what is the smallest angle from the horizontal taht will cause the egg to slide?
Here's my drawing:
(http://i51.photobucket.com/albums/f362/BeamsBox/PhysicsEgg.jpg)
Here's my work:
x, implying that it's along the x-axis
Fs is static friction force
Fg is the force of gravity
m = mass
a = acceleration
(mue) = .4, the coefficient of friction
Fs(max) = (mue)FN Equation 1
FN = mg
Substitute to get:
Fs(max) = (mue)(mg) Equation 2
F(net)x = max
Fs - Fgx = max
Since it is stationary, a = 0
Fs - Fgx = m(0)
Fs = Fgx Equation 3
Now substitute (mue)(mg) for Fs,
(mue)(mg) = Fgx, and substitute the Fgx from the diagram,
(mue)(mg) = Fgsin(theta), and substitute mg for Fg
(mue)(mg) = (mg)sin(theta), mg cancels out
(mue) = sin(theta), and substitute the known .4 for (mue),
.4 = sin(theta)
theta = sin-1(.4)
theta = about 23.6 degrees
The book says 2... any ideas on where I may have gone wrong?
Given the coefficient of static friction (.4) between an egg and a pan, what is the smallest angle from the horizontal taht will cause the egg to slide?
Here's my drawing:
(http://i51.photobucket.com/albums/f362/BeamsBox/PhysicsEgg.jpg)
Here's my work:
x, implying that it's along the x-axis
Fs is static friction force
Fg is the force of gravity
m = mass
a = acceleration
(mue) = .4, the coefficient of friction
Fs(max) = (mue)FN Equation 1
FN = mg
Substitute to get:
Fs(max) = (mue)(mg) Equation 2
F(net)x = max
Fs - Fgx = max
Since it is stationary, a = 0
Fs - Fgx = m(0)
Fs = Fgx Equation 3
Now substitute (mue)(mg) for Fs,
(mue)(mg) = Fgx, and substitute the Fgx from the diagram,
(mue)(mg) = Fgsin(theta), and substitute mg for Fg
(mue)(mg) = (mg)sin(theta), mg cancels out
(mue) = sin(theta), and substitute the known .4 for (mue),
.4 = sin(theta)
theta = sin-1(.4)
theta = about 23.6 degrees
The book says 2... any ideas on where I may have gone wrong?
Last edited: