67 kg tree surgeon rides a \"cherry picker\" lift (Fig. 5-34) to reach the upper

Question

67 kg tree surgeon rides a "cherry picker" lift (Fig. 5-34) to reach the upper branches of a tree.

(a) What force does the bucket of the lift exert on the surgeon when the bucket is at rest?
N

(b) What is the force when the bucket is moving upward at a steady 3.0 m/s?
N

(c) What is the force when the bucket is moving downward at a steady 3.0 m/s?
N

(d) What is the force when the bucket is accelerating upward at 2.4 m/s2?
N

(e) What is the force when the bucket is accelerating downward at 2.4 m/s2?
N

Explanation / Answer

a) system is at rest and in equilibrium.

Fnet = N - mg = 0

N = 67 x 9.81 = 657.27 N

b) system is moving at steady speed, so acceleration is zero

so system is in equilibrium.

N =657.27 N

c) system is moving at steady speed, so acceleration is zero

so system is in equilibrium.

N =657.27 N

d) using newton's 2nd law in vertical upward direction,

N - mg = ma

N = m(g + a) = 67 (9.81 + 2.4) = 818.07 N

e) using newton's 2nd law in vertical downward direction,

mg - N = ma

N = m(g - a) = 67 (9.81 - 2.4) = 496.47N

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