Climber hanging from a cliff and acceleration

[cardhouse42]

Homework Statement

A 75 kg climber finds himself dangling over the edge of an ice cliff. Fortunately, he's roped to a 920 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.3×10^−2 . What is his acceleration? Neglect the rope's mass.

Homework Equations

Tension(rock)=tension(climber)=T
acceleration(rock)=acceleration(climber)=a

Climber(y):T-m(c)*g=m(c)*a
Rock(x):T+f(k)=m(r)*a
Rock(y):N-m(r)*g=0

The Attempt at a Solution

T=m(c)*a+m(c)*g
f(k)=mu(k)*N
N=m(r)*g

m(c)*a+m(c)*g+mu(k)*m(r)*g=m(r)*a
a=[m(c)*g+mu(k)*m(r)*g]/[m(r)-m(c)]
a=1.44 m/s^2

The computer is telling me that I have the wrong answer. Do I need to take the 51 m into account somehow? Any help would be greatly appreciated!

 

[giggidygigg]

friction opposes relative motion. the rock wants to move in the direction of the tension of the rope, but the friction opposes it, which means its direction is opposite.

 

[cardhouse42]

friction opposes relative motion. the rock wants to move in the direction of the tension of the rope, but the friction opposes it, which means its direction is opposite.

Thanks! That gave me the right answer :)

 

[chaoseverlasting]

Write out the individual equations of motion for the rock and the man, add them to eliminate T and solve for a. You do not need to take the length of the rope into account.