Question

A bridge, constructed of 11 beams of equal length and negligible mass, supports an object of mass as shown. Real bridges of this sort have steel rockers at the ends, these assure that the support forces on the bridge are vertical even when it expands or contracts thermally.

Find F_P, the vertical force that pier P exerts on the left end of the bridge.

Answer 1

This can be solved by calculating the total moment of the bridge which is zero. Taking point Q as the pivot of the moment and the clockwise moment is positive, we could solve the problem now

Moment = 0
F_P * 3L - M * 2L = 0
F_P - M * 2 / 3 = 0
F_P = 2M/3.

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Answer 2

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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Answer details

Grade: High School

Subject: Physics

Chapter: Moment of Force