Question

a uniform beam of length l and mass mb is supported by two pillars located l/3 from either end, as shown in the figure. a duck of mass md stands on one end. a scale is placed under each pillar. the entire system is in equilibrium.

Answer 1

When the system is in equilibrium, the sum of the moment about a point is

zero.

$Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar \ is \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}$

Reasons:

Length of the beam = l

Mass of the beam = $m_B$

$Location \ of \ the \ two \ pillars = \dfrac{l}{3} \ from \ either \ end$

Mass of the duck = $m_D$

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 = $m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3}$

Anticlockwise moment = $m_B \times g \times \dfrac{l}{6}$

At equilibrium, clockwise moment = Anticlockwise moment

Therefore;

$m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6}$

$F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}$

$F = \dfrac{m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}}{\dfrac{l}{3} } = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}$

$Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar, \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}$

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