How Much Force Do Pillars Exert with a Parked Car on a Bridge?

  • Thread starter Thread starter chamonix
  • Start date Start date
  • Tags Tags
    Bridge Rotational
AI Thread Summary
The discussion centers on calculating the force exerted by two pillars supporting a bridge with a parked car. The bridge is 20 m long and weighs 4.00x10^5 N, while the car weighs 1.96x10^4 N and is parked 8 m from one end. The initial calculation provided is incorrect due to misidentified distances from the pivot point and the normal force of the pillars. Participants emphasize the importance of accurately determining distances and suggest drawing a diagram for clarity. Correct calculations for each pillar will yield different force values.
chamonix
Messages
21
Reaction score
0

Homework Statement


A uniform bridge 20 m long and weighing 4.00x10^5N is supported by two pillars located 3 m from each end. If a 1.96x10^4 N car is parked 8 m from one end of the brige, how much force does each pillar exert?


Homework Equations


(Fgb*L/2+Fgp*d)/L(sin theta).


The Attempt at a Solution


4x10^5*10+1.96x10^4*8/20(sin 90)=207840
That's what I got as an answer. I don't know if this is correct though. Any help is appreciated. Thank you.
 
Physics news on Phys.org
No you don't have all of your distances right. The car is 8 m from one end, but how far is it from the pivot point you've chosen (one of the pillars)? Also, the distance of the normal force of the pillar to the pivot point is not 20 m. Draw a diagram and label everything carefully.

Repeat for the other pillar, the answer will be different.
 
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