Question

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 23.5 cm.
The bat has a mass of 1.30 kg and has a center of mass that is 70.0 cm from the small end of the bat.(a) Find the magnitude of the force exerted by the hand.(b) Find the direction of the force exerted by the hand.(c) Find the magnitude of the force exerted by the shoulder.(d) Find the direction of the force exerted by the shoulder.

Answer 1

Answer:

a) Fh = 25.23 N

b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.

c) Fs = 37.97 N

d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.

Explanation:

Given data

Distance from the small end of the bat to the shoulder: d = 23.5 cm

Distance from the small end of the bat to the center of mass: r = 70.0 cm

Mass of the bat: m = 1.30 Kg

This situation can be seen in the pic.

a) We can apply

∑τA = 0   ⇒  + (23.5cm)*Fh - (70 - 23.5)cm*m*g = 0

⇒   + (23.5cm)*Fh - (70 - 23.5)cm*(1.30 Kg)*(9.8 m/s²) = 0

⇒   Fh = 25.23 N (↓)

b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.

c) We apply

∑Fy = 0  (↑)

⇒      - Fh + Fs - m*g = 0

⇒     Fs = Fh + m*g

⇒     Fs = 25.23 N + (1.30 Kg)*(9.8 m/s²)

⇒     Fs = 37.97 N (↑)

d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.