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A 89.3 kg man sits on the stern of a 5.8 m
long boat. The prow of the boat touches the
pier, but the boat isn’t tied. The man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 4.17 m away from the pier.
Assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
Answer in units of kg

Respuesta :

MathPhys

Answer:

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

Ver imagen MathPhys
snehashish65

The required mass of boat is 34.9 kg.

The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.

Given data:

The mass of man is, m = 89.3 kg.

The length of boat is, L = 5.8 m.

The distance away from the pier is, d = 4.17 m.

Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m d + M (d + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)

m L + M (L/2) = m d + M (d + L/2)

m L + M (L/2) = m d + M d + M (L/2)

Further solving as,

m L = m d + M d

m L − m d = M d

m (L − x) = M x

M = m (L − x) / x

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

Thus, we can conclude that the required mass of boat is 34.9 kg.

Learn more about the center of mass here:

https://brainly.com/question/8662931

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