What is the vertical force exerted by pier P on the left end of the bridge?

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The discussion revolves around determining the vertical force exerted by pier P on the left end of a bridge, which is supported by multiple beams and an object of mass M. The problem involves concepts of forces, torques, and gravitational effects on the structure.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to relate the torque due to the mass and the weight of the bridge to find the force exerted by pier P. Some participants question the correctness of the calculations, particularly regarding factors that may have been omitted.

Discussion Status

Participants are actively engaging with the problem, checking each other's calculations, and clarifying points of confusion. There is a mix of attempts to solve the problem and requests for explanations regarding specific steps in the reasoning.

Contextual Notes

There is a mention of changing the question, which some participants caution against, as it may disrupt the coherence of the discussion. The original poster expresses frustration with previous attempts to solve the problem.

vorcil
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FORCES On a bridge

A bridge, constructed of 11 beams of equal length L
and negligible mass, supports an object of mass M


as shown. (Intro 1 figure) Real bridges of this sort have steel rockers at the ends (check one out sometime), these assure that the support forces on the bridge are vertical even when it expands or contracts thermally.

Throughout this problem, use g for the magnitude of the acceleration due to gravity.

MFS_st_1_a.jpg



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


My attempt:
I know the mass exerts a clockwise torque on the left end
and that their is also the force of the whole bridge on the bridge due to gravity

so Fp = Torqueduetomass + the whole weight force /2 (because it's halved? due to theother end of the bridge helping out)
 
Last edited:
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vorcil said:
Substituting in (1) for h from (2), I from (3), and w from (4):
(mv^2 + 2mv^2 / 3) = mgs sin(a)
Looks like you dropped a factor of 2.
 
Doc Al said:
Looks like you dropped a factor of 2.


wait can you please explain?

(mv^2 + 2mv^2/3) /2 is correct?
 
vorcil said:
(mv^2 + 2mv^2/3) /2 is correct?
Yes, that is correct. (Don't lose that 2 in the denominator.)
 
Doc Al said:
Yes, that is correct. (Don't lose that 2 in the denominator.)

thanks, I've changed the question now this is one I've had ages to do but keep getting it wrong
 
vorcil said:
thanks, i've changed the question now this is one I've had ages to do but keep getting it wrong
Don't go back and change the question. That turns the thread into gibberish.
 

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