Force Exerted by Road on an Accelerating Car

AI Thread Summary
To determine the force exerted by the road on the SUV, the acceleration of the vehicle is calculated to be approximately 2.98 m/s². Using Newton's second law, the horizontal force (Fx) is found to be 6854.61 N, while the vertical force (Fy) due to gravity is 22540 N. The user is uncertain about how to combine these forces to find the total force exerted by the road. They mention a method involving the square root of the sum of the squares of Fx and Fy, but this approach did not yield the correct answer. Clarification on the correct method to find the resultant force is requested.
yoshi824
Messages
1
Reaction score
0

Homework Statement



A huge SUV of mass 2300 kg can accelerate from 0 to 60 mi/h in 9.0 s. What force does the road exert on the SUV?

V0 = 0
V= 60mph or 26.8224m/s
M= 2300kg
t=9s

Homework Equations



F=ma

The Attempt at a Solution


a= (26.8224) / 9 = 2.98
Fx = ma = 2.98*2300 = 6854.61 N

Fy = m*g = 2300*9.8 = 22540 N

So those are the two forces that I was able to calculate but I am not sure how I find the force exerted by the road on the car. Any help would be greatly appreciated! Thanks!

other info: I read somewhere that the force would be sqrt (Fx2 + Fy2) but that didn't give me the right answer if my above work is correct.
 
Physics news on Phys.org
If you made a mistake I would like to know what it is. Your work looks right to me.
 
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