. A rope 2 metres long rests on a table where the coefficient of friction is 0.5. What minimum length of rope can stay on the table before the whole rope will slip off?

Question

a) 133 cm
b) 80 cm
c) 200 cm
d) Cannot determine without more information.

The answer is A but i can't figure out how to do it.

Answer 1

force friction=mu*lengthdensity*L*g
force hanging over=lengthdensity*(2-L)*g

set them equal
mu*L*sigma*g=sigma*(2-L)g
sigma*g(.5L-2+L)=0
1.5L=2
L=4/3 meter or answer a.

Answer 2

length on table = x
length over side = 2-x

mass/meter length = m

mass on table = m x
max friction force =.5 m g x

mass over the side = m(2-x)
weight over the side = m g(2-x)

if slips when
.5 m g x = m g (2-x)

x = 2 (2-x)
3 x = 4
x = 4/3 = 1.33 meter = 133 cm