Kinetic Friction!

  • Thread starter orange03
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Homework Statement



A 75 kg climber finds himself dangling over the edge of an ice cliff, as shown in the figure below. Fortunately, he's roped to a 980 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.5×10−2. What is his acceleration, and how much time does he have before the rock goes over the edge? Neglect the rope's mass. What is his acceleration? How much time does he have before the rock goes over the edge?




Homework Equations


F=ma


The Attempt at a Solution



equation of the climber:
Ft-mg=may

equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N. I plugged this back into the climber's equation and got -2.76 m/s^2 but this answer is wrong. Help please!
 

Answers and Replies

  • #2
1,662
130
equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N.

The equation of the rock is wrong
The tension is not the same as friction because the rock moves with acceleration.
Use [tex]\sum F[/tex] = ma to get the equation for the rock
 

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