Rock climber force diagram and acceleration

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The discussion focuses on calculating the acceleration of a 70kg climber and a 940kg rock in a frictionless scenario. The climber's downward force is determined using F = mg, resulting in a force of 686.7 N. By applying Newton's second law (F = ma), the acceleration of both the climber and the rock can be derived as 0.073 m/s². Additionally, the time it takes for the rock to fall over the edge is calculated using the formula a = distance/time², allowing for the determination of time based on known acceleration and distance.

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A 70kg climber is dangling over the edge of an ice cliff. He is roped to a 940kg rock located 51 m from the edge. The ice is frictionless, and the climber starts to accelerate downward.

a) How would you find the acceleration of the climber?

Would you use the force diagrams for the rock and the climber and subtract the two to find the acceleration in the case?


b) How long does it take the rock to go over the edge?

Much thanks
 
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a) the force on the rock from the climber is 70 x 9.81 (F = mg). And we know that F = ma. So 70x 9.8 = 940 x a. Solve for a. (the acceleration of the rock is = to the acceleration of the climber.

b) a = distance/time squared. You know a, and distance. Solve for t.
 

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