When the system is in equilibrium, the sum of the moment about a point is
zero.
[tex]Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar \ is \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}[/tex]
Reasons:
Length of the beam = l
Mass of the beam = [tex]m_B[/tex]
[tex]Location \ of \ the \ two \ pillars = \dfrac{l}{3} \ from \ either \ end[/tex]
Mass of the duck = [tex]m_D[/tex]
Required:
Force shown by the scale under the right pillar.
Solution:
The location of the duck = On the left end of the beam
When the system is in equilibrium, we have; ∑M = 0
Taking moment about the left pillar, we get;
Clockwise moment = [tex]m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3}[/tex]
Anticlockwise moment = [tex]m_B \times g \times \dfrac{l}{6}[/tex]
At equilibrium, clockwise moment = Anticlockwise moment
Therefore;
[tex]m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6}[/tex]
[tex]F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}[/tex]
[tex]F = \dfrac{m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}}{\dfrac{l}{3} } = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}[/tex]
[tex]Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar, \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}[/tex]
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