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A 67 kg man stands at the front end of a uniform boat with a mass of 179 kg and a length of [tex]L = 2.5 \, \text{m}[/tex]. Assume there is no friction or drag between the boat and the water.

(a) What is the location of the center of mass?

Answer :

Final answer:

The location of the center of mass of the boat with the man standing at the front end is undefined.

Explanation:

The center of mass is the point at which the entire mass of an object can be considered to be concentrated. To find the center of mass of the boat with the man standing at the front end, we can use the concept of torque. The torque exerted by the man and the boat around the center of mass should balance out to zero for the system to be in equilibrium.

The torque exerted by the man is given by the product of his weight (m1*g) and the distance between his position and the center of mass of the boat (d1), which is equal to half the length of the boat (L/2). The torque exerted by the boat itself is given by the product of its weight (m2*g) and the distance between its center of mass and the center of mass of the system (d2), which is equal to half the length of the boat (L/2) as well.

Setting up the equation for torque balance:

m1*g*d1 = m2*g*d2

Substituting the given values:

(67 kg)*(9.8 m/s^2)*(L/2) = (179 kg)*(9.8 m/s^2)*(L/2)

Canceling out common factors:

67(L/2) = 179(L/2)

Simplifying the equation:

67 = 179

This is not a valid equation, which means the given system cannot be in equilibrium. Therefore, the center of mass of the boat with the man standing at the front end is undefined.

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