Answer:
[tex]m_b=278.73\ kg[/tex]
Explanation:
Center of Gravity
It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below
[tex]\displaystyle x_c=\frac{\sum x_im_i}{\sum m_i}[/tex]
Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at
[tex]\displaystyle x_c=\frac{(4.9)(90.4)+(2.45)(m_b)}{90.4+m_b}[/tex]
[tex]\displaystyle x_c=\frac{442.96+2.45m_b}{90.4+m_b}[/tex]
When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance
[tex]x_b=1.2+2.45=3.65\ m[/tex]
The man is located at
[tex]x_m=1.2\ m[/tex]
The center of gravity is computed now as
[tex]\displaystyle x_c=\frac{(1.2)(90.4)+(3.65)(m_b)}{90.4+m_b}[/tex]
[tex]\displaystyle x_c=\frac{108.48+3.65m_b}{90.4+m_b}[/tex]
Both centers of gravity are equal, thus
[tex]\displaystyle \frac{442.96+2.45m_b}{90.4+m_b}=\frac{108.48+3.65m_b}{90.4+m_b}[/tex]
Simplifying
[tex]442.96+2.45m_b=108.48+3.65m_b[/tex]
Rearranging
[tex]1.2m_b=334.48[/tex]
Thus
[tex]\boxed{m_b=278.73\ kg}[/tex]
