What Force Does a Concrete Support Exert at One End of a Bridge?

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SUMMARY

The discussion revolves around calculating the force exerted by a concrete support at one end of a bridge when a hiker weighing 986 N stops one-fifth of the way along the bridge, which weighs 3550 N. The problem requires applying torque equilibrium principles, specifically using the equation T = F x d. The correct approach involves summing torques about the right end of the bridge to account for both the hiker's weight and the bridge's weight, leading to the determination of the force at the near support.

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  • Understanding of torque and equilibrium principles in physics
  • Familiarity with the concept of weight and gravitational force
  • Ability to apply the equation T = F x d in practical scenarios
  • Knowledge of how to sum forces and torques in a static system
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  • Learn how to calculate torques about different pivot points
  • Explore examples of similar problems involving forces on beams and bridges
  • Investigate the effects of varying load positions on support forces
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Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as engineers involved in structural analysis and design of bridges.

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Homework Statement



A hiker, who weighs 986 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3550 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge at the near end

Weight of Person = 986N
Weight of Bridge = 3550N
d = 1/5th of Bridge length


Homework Equations


This is a torque question, so the equation is T = F x d
i know that the force is the person times gravity, but that's already done.
However i don't know the length of the bridge, I've tried dividing the mans mass by 5 and adding that to the bridge's weight, but to no prevail. i really need some help here guys


The Attempt at a Solution



no really attempts
T = 986 x 0.2
T = 197.2
T = 197.2 + 3550
T = 3747.2
 
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Hypnos_16 said:

Homework Statement



A hiker, who weighs 986 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3550 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge at the near end

Weight of Person = 986N
Weight of Bridge = 3550N
d = 1/5th of Bridge length


Homework Equations


This is a torque question, so the equation is T = F x d
i know that the force is the person times gravity, but that's already done.
However i don't know the length of the bridge, I've tried dividing the mans mass by 5 and adding that to the bridge's weight, but to no prevail. i really need some help here guys


The Attempt at a Solution



no really attempts
T = 986 x 0.2
T = 197.2
T = 197.2 + 3550
T = 3747.2
Tis is a problem in equilibrium where the net sum of the torques from each force about any point must equal 0. Try summing torques about the right end of the bridge. You will have torques produced by both the person's weight force, the bridge's weight force, and the left support unknown force. Where does the resultant force of the bridge's weight act??
 

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