Sliding egg I think the books wrong, could someone please verify?

1. Jun 17, 2010

Beamsbox

***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)

The book says 2... any ideas on where I may have gone wrong?

Last edited: Jun 17, 2010
2. Jun 17, 2010

D H

Staff Emeritus
The magnitude of the normal force is not equal to mg.

3. Jun 17, 2010

kuruman

To elaborate on DH's remark. If the egg is just about to slide, it is still at rest and the sum of all the forces on it must be zero. Look at your drawing. For the sum of the three vectors to be zero, the magnitudes of the vectors must form a closed triangle. Fg is the hypotenuse and Fn one of the right sides. Can they be equal? What is it that must be equal instead?

4. Jun 17, 2010

Beamsbox

FN = Fgcos(theta)... that seems to make more sense.

5. Jun 17, 2010

kuruman

Can you finish the problem now?

6. Jun 19, 2010

Beamsbox

Ya, I figured it out... finally thanks again.