Babe Ruth steps to the plate and casually points to left center field to indicate the location of his next home run. The mighty Babe holds his bat across his shoulder, with one hand holding the small end of the bat. The bat is horizontal, and the distance from the small end of the bat to the shoulder is 24.5 cm . The bat has a mass of 1.50 kg and has a center of mass that is 68.0 cm from the small end of the bat.

find the magnitudeand direction of the force exerted bya) the handb) the shoulder.

ANSWER: (a) (-26.1N)y , (b) 40.82Ny

Explanation:

given the distance from the small end of the bat to shoulder=24.5cm

bat mass = m = 1.50kg

center of mass distance from the bat = 68cm

place the center of rotation at the shoulder

the y - axis is upward

then force exerted by hand = Στ= 24.5cm)Fh -( 68cm -24.5cm)mg = 0

  Fh   =(43.5cm)(1.5kg)(9.81m/s2)/24.5cm

         =26.1N

Fh= (-26.1N)y

force exerted on the shoulder

ΣFy = -Fh+Fs -mg =0

Fs= Fh +mg = 26.1N +1.5kg(9.81m/s2) = 40.82N

Fs= 40.82Ny

i think the answer would be reserve.