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 22.5 cm . The bat has a mass of 1.10 kg and has a center of mass that is 67.0 cm from the small end of the bat.
a) Find the magnitude of the force exerted by the hand?
b)Find the magnitude of the force exerted by the shoulder.
and their directions
Thanks!
hand at x = 0
shoulder at x = .225meters
bat at .67meters
hand (moments about shoulder = 0)
1.1 (9.8)(.67 - .225) = Fh(.225)
Fh is down
shoulder (moments about hand = 0)
1.1(9.8)(.67) = Fs (22.5)
Fs is up
To find the magnitude of the force exerted by the hand and the shoulder, we can use the principle of torque and equilibrium.
a) Finding the magnitude of the force exerted by the hand:
- Given that the distance from the small end of the bat to the shoulder (L) is 22.5 cm or 0.225 m.
- The center of mass of the bat is 67.0 cm from the small end or 0.67 m.
- The mass of the bat is 1.10 kg.
Since the bat is in equilibrium (not rotating), the sum of the torques acting on it must be zero. Torque is given by the equation:
τ = r × F,
where τ is the torque, r is the displacement vector from the axis of rotation (shoulder) to the point of force application (hand), and F is the force applied.
- The torque due to the force exerted by the hand is countered by the torque due to the force exerted by the shoulder.
- The torque due to the hand is given by τ_hand = r_hand × F_hand, where r_hand = 0.225 m and F_hand is the force exerted by the hand.
- The torque due to the shoulder is given by τ_shoulder = r_shoulder × F_shoulder, where r_shoulder = 0.67 m and F_shoulder is the force exerted by the shoulder.
Since the bat is not rotating, τ_hand = τ_shoulder. Therefore:
r_hand × F_hand = r_shoulder × F_shoulder
0.225 m × F_hand = 0.67 m × F_shoulder
Divide both sides by 0.225 m:
F_hand = (0.67 m / 0.225 m) × F_shoulder
F_hand = 2.98 × F_shoulder
So, the magnitude of the force exerted by the hand is approximately 2.98 times larger than the magnitude of the force exerted by the shoulder.
b) Finding the magnitude of the force exerted by the shoulder:
We can use the equation derived in part a:
F_hand = 2.98 × F_shoulder
Since we know the mass of the bat (1.10 kg) and the acceleration due to gravity (9.8 m/s²), we can calculate the weight of the bat using the equation:
Weight = mass × acceleration due to gravity
Weight = 1.10 kg × 9.8 m/s²
The weight of the bat is given by the formula:
Weight = Force_shoulder - Force_hand
So, we can rewrite the equation from part a as:
1.10 kg × 9.8 m/s² = F_shoulder - 2.98 × F_shoulder
Simplifying and solving for F_shoulder:
(1.10 kg × 9.8 m/s²) = (1 - 2.98) × F_shoulder
(1.10 kg × 9.8 m/s²) = (-1.98) × F_shoulder
Divide both sides by -1.98:
F_shoulder = -(1.10 kg × 9.8 m/s²) / -1.98
So, the magnitude of the force exerted by the shoulder is approximately 5.49 N, directed opposite to the force exerted by the hand.
To find the magnitude of the force exerted by the hand, we can use the principle of moments and equate the torques about the center of mass of the bat.
a) The torque exerted by the hand is given by the equation:
T_hand = F_hand * d_hand
Where T_hand is the torque exerted by the hand, F_hand is the force exerted by the hand, and d_hand is the distance from the center of mass to the hand.
Since the bat is at rest and in equilibrium, the sum of all torques must be zero. Therefore, we have:
T_hand + T_shoulder = 0
Now, let's substitute the known values into the equation:
F_hand * d_hand + F_shoulder * d_shoulder = 0
We know that the distance from the center of mass to the hand, d_hand, is given as 22.5 cm, and the distance from the center of mass to the shoulder, d_shoulder, is given as 67.0 cm. Let's convert these distances to meters:
d_hand = 0.225 m
d_shoulder = 0.670 m
Substituting these values into the equation, we have:
F_hand * 0.225 + F_shoulder * 0.670 = 0
Since Babe Ruth is only holding the bat with one hand, the force exerted by the shoulder, F_shoulder, is negligible compared to the force exerted by the hand. Therefore, we can ignore the term involving F_shoulder, and the equation becomes:
F_hand * 0.225 = 0
Simplifying further, we get:
F_hand = 0
This means that the magnitude of the force exerted by the hand is zero.
b) Since the magnitude of the force exerted by the hand is zero, there is no force exerted by the shoulder. Therefore, the magnitude of the force exerted by the shoulder is also zero.
In summary:
a) The magnitude of the force exerted by the hand is zero.
b) The magnitude of the force exerted by the shoulder is zero.