The vertical force that pier P exerts on the left end of the bridge,
Fp = ²/₃ Mg
The vertical force that pier Q exerts on the right end of the bridge,
Fq = ¹/₃ Mg
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Further explanation
Let's recall Moment of Force as follows:
\(\boxed{\tau = F d}\)
where:
τ = moment of force ( Nm )
F = magnitude of force ( N )
d = perpendicular distance between force and pivot ( m )
Let us now tackle the problem !
Given:
mass of the object = M
gravitational acceleration = g
Asked:
vertical force at pier P = Fp = ?
vertical force at pier Q = Fq = ?
Solution:
If the bridge is in equilibrium position , then :
\(\texttt{Total Clockwise Moment at Q = Total Anticlockwise Moment at Q }\)
\(F_p \times PQ = w \times QR\)
\(F_p \times 3L = Mg \times 2L\)
\(F_p = 2MgL \div (3L)\)
\(\boxed{F_p = \frac{2}{3}Mg}\)
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\(\texttt{Total Upward Force = Total Downward Force }\)
\(F_p + F_q = Mg\)
\(\frac{2}{3}Mg + F_q = Mg\)
\(F_q = Mg - \frac{2}{3}Mg\)
\(\boxed{F_q = \frac{1}{3}Mg}\)
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
- Newton's Law of Motion: https://brainly.com/question/10431582
- Example of Newton's Law: https://brainly.com/question/498822
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Answer details
Grade: High School
Subject: Physics
Chapter: Moment of Force
