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

  • Thread starter Thread starter Hypnos_16
  • Start date Start date
  • Tags Tags
    Bridge Support
AI Thread Summary
The discussion focuses on calculating the force exerted by a concrete support on a bridge when a hiker stops one-fifth of the way along it. The hiker weighs 986 N and the bridge weighs 3550 N, creating a torque problem that requires understanding of equilibrium. Participants suggest using the torque equation T = F x d and recommend summing torques about the right end of the bridge to find the unknown force. The challenge lies in determining the effective length of the bridge and the point where the bridge's weight acts. Overall, the problem emphasizes the need for applying principles of static equilibrium to solve for the forces involved.
Hypnos_16
Messages
148
Reaction score
1

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
 
Physics news on Phys.org
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??
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top