Question

1) A reaction board of length ` = 2 m and mass mb = 10 kg is used to measure the center of mass of a person. On one end of the board there is a pivot, on the other a scale. A person with mass m = 60 kg lies with their feet at the pivot. The scale registers a mass of ms = 30 kg.

(i) Draw a force diagram of the reaction board and label all forces and relevant distances.

(ii) How far is the center of mass of the person from the person’s foot? Express your answer in terms of `, ms, m, and mb. Then substitute numbers to find the numerical answer: you should get a distance d = 0.83 m.

(iii) What is the force at the person’s foot? Express your answer in terms of m, mb, ms, and g. Then substitute numbers to find the numerical answer: you should get a force Ff = 392 N

Answer 1

(a) Here is the free body diagram

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(b) taking moment about pivot, we have

Let d be the distance from center of mass to feet

then

md + m_bL/2 - m_sL = 0

d = m_sL - 0.5m_bL / m

Now, put in the values

d = 30 * 2 - 0.5 * 10 * 2 / 60

d = 50 / 60

d = 0.83 m

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(c) just equate all the forces in vertical direction, note that force at feet ( pivot) will act upward.

then

upward forces = downward force

F_f + m_sg = m_bg + mg

F_f = m_bg + mg - m_sg

F_f = ( m_b + m - m_s ) g

now just put the values

F_f = ( 10 + 60 - 30) * 9.8

F_f = 392 N