Speed bump question - Newton's laws

AI Thread Summary
When driving over a speed bump at 8 m/s, the sensation of being lifted from the seat is due to inertia, as the body tends to maintain its state of motion while the car changes direction. The force exerted by the seat on the driver is influenced by the centripetal acceleration experienced during the bump. Using the formula F = mv²/r, the weight of the driver (700 N) is reduced by the centripetal force, leading to a decrease in the normal force from the seat. The driver does not float off the seat but experiences a temporary reduction in the force pressing them down. Ultimately, the seat does not exert a constant 700 N during the bump due to this dynamic change in forces.
feveroffate
Messages
8
Reaction score
0
1.
You're driving down the road at 8 m/s and go over a speed bump. When you go over the bump, you feel lifted from your seat. Explain why.


I assumed it was because of inertia, but I'm not sure.

You weigh 700 N. If it is approximated that the speed bump is shaped like an arc of a circle, with radius of 10 m, find the magnitude of force the seats exerts on you as you go over the speed bump.

If F = mv^2/r, then would you just plug everything in directly?
 
Physics news on Phys.org
The seat normally pushes upward on you with 700 N (because you push down on it with 700 N weight). While in circular motion, though, you experience a centrifugal force upward that reduces the 700 N.
 
So while in circular motion, you aren't pushing down on the seat with 700 N so you are lifted up?
 
It is unlikely you will float up off the seat - that would take a terrific speed - but the force pressing you against the springs of the seat will be reduced so the spring compression will be less and you will go up a little bit.
 
But would the seat still exert 700 N upward? I mean, would I just plug in the numbers into the equation for the second part?
 
700 - mv^2/r
 
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