Rock climber force diagram and acceleration

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To determine the acceleration of a 70kg climber dangling over an ice cliff, one can use force diagrams for both the climber and the 940kg rock, applying Newton's second law (F = ma). The force exerted by the climber is calculated as 70kg multiplied by 9.81m/s², which equals 686.7N. This force is balanced against the rock's mass to find the shared acceleration, leading to the equation 70 x 9.81 = 940 x a, allowing for the calculation of acceleration. To find the time it takes for the rock to go over the edge, the formula a = distance/time² can be rearranged to solve for time, using the known acceleration and distance of 51m. This analysis provides a clear method for calculating both the climber's acceleration and the time for the rock's descent.
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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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