Solve Kinetic Friction Homework: Acceleration & Time Before Rock Goes Over Edge

AI Thread Summary
The discussion centers on a physics homework problem involving a climber and a rock on an ice cliff, requiring the calculation of the climber's acceleration and the time before the rock falls over the edge. The climber's equation is set up using the force of tension and gravitational force, while the rock's equation involves the frictional force. A participant calculated the tension in the rope but arrived at an incorrect acceleration of -2.76 m/s². It is suggested that the equations need to be revised, particularly for the rock, emphasizing the need to apply the sum of forces to derive the correct relationship. Accurate calculations are essential for determining both the climber's acceleration and the time before the rock falls.
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!
 
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orange03 said:
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 \sum F = ma to get the equation for the rock
 
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