Answer:
Case A) tau_net = -243.36 N m, case B) tau_net = 783.36 N / m, tau_net = -63.36 N m, case C) tau _net = - 963.36 N m,
Explanation:
For this exercise we use Newton's relation for rotation
Σ τ = I α
In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N
we will assume that the counterclockwise turns are positive.
case a
tau_net = m g x - F x2
tau_nett = -28.8 9.8 1.5 + 180 1
tau_net = -243.36 N m
in this case the man's force is downward and the system rotates clockwise
case b
2 force clockwise, the direction of
the force is up
tau_nett = -28.8 9.8 1.5 - 180 2
tau_net = 783.36 N / m
in case the force is applied upwards
3) counterclockwise
tau_nett = -28.8 9.8 1.5 + 180 2
tau_net = -63.36 N m
system rotates clockwise
case c
2 schedule
tau_nett = -28.8 9.8 1.5 - 180 3
tau _net = - 963.36 N m
3 counterclockwise
tau_nett = -28.8 9.8 1.5 + 180 3
tau_net = 116.64 Nm
the sitam rotated counterclockwise
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