Answer
(a) Please refer to the free-body diagram.
(b) $F_x = 77~N$
(c) $F_x = 12\times ~w$
Work Step by Step
(a) Please refer to the free-body diagram.
(b) We can find the mass of the hand.
$m = 0.0125\times 52~kg = 0.65~kg$
We can find the speed $v$.
$v = \frac{2.0\times 2\pi ~r}{t} = \frac{(2.0)(2\pi)(0.75~m)}{1~s}$
$v = 3.0~\pi~m/s$
We can find the horizontal force $F_x$ exerted by the wrist on the hand.
$F_x = \frac{mv^2}{r} = \frac{(0.65~kg)(3.0~\pi~m/s)^2}{0.75~m}$
$F_x = 77~N$
(c) We can express the force as a multiple of the weight of the hand $w$ (which is equal to $mg$).
$F_x = \frac{77~N}{(0.65~kg)(9.80~m/s^2)} = 12\times ~w$
