What is the mass of this car based on given forces?

  • Thread starter Thread starter Ace.
  • Start date Start date
  • Tags Tags
    Car Mass
AI Thread Summary
The discussion revolves around calculating the mass of a car based on the forces acting on it. Each wheel exerts a downward force of 4.0 x 10^3 N, totaling 16,000 N downward. The net horizontal force is determined to be 2,000 N west after accounting for driving and frictional forces. There is uncertainty regarding whether the normal force is simply the total downward force from the wheels and how horizontal forces affect vertical weight distribution. The conversation highlights the complexity of balancing forces in a dynamic scenario, suggesting that additional information may be needed to accurately determine the car's mass.
Ace.
Messages
52
Reaction score
0

Homework Statement



Each of the four wheels of a car pushes on the road with a force of 4.0 x 103 N [down]. The driving force on the car is 8.0 x 103 N [west]. The frictional resistance on the car is 6.0 x 103 N [east]. Calculate the mass of the car.

Homework Equations



F = ma
Fnet = F1 + F2


The Attempt at a Solution



Fnet = F1 + F2
Fnet = FA - Ff
= 8.0 x 103 - 6.0 x 103
= 2000 N [west]​

Total force of all four wheels:
(4.0 x 103 N [down]) x 4
= 16000 N [down]​

FN = 16000 N[up]
But this is only part of the Normal Force, because the rest is coming from the weight of the car.

It seems to me that there isn't enough info to find mass?
 
Physics news on Phys.org
Wait, is the normal force simply the force of all the wheels ?
 
Ace. said:

Homework Statement



Each of the four wheels of a car pushes on the road with a force of 4.0 x 103 N [down]. The driving force on the car is 8.0 x 103 N [west]. The frictional resistance on the car is 6.0 x 103 N [east]. Calculate the mass of the car.

Homework Equations



F = ma
Fnet = F1 + F2


The Attempt at a Solution



Fnet = F1 + F2
Fnet = FA - Ff
= 8.0 x 103 - 6.0 x 103
= 2000 N [west]​

Total force of all four wheels:
(4.0 x 103 N [down]) x 4
= 16000 N [down]​

FN = 16000 N[up]
But this is only part of the Normal Force, because the rest is coming from the weight of the car.

It seems to me that there isn't enough info to find mass?

Ace. said:
Wait, is the normal force simply the force of all the wheels ?

That would be my answer. The horizontal movement forces in this case would not seem to affect the vertical weight.
 
Actually, it looks from the sum of the horizontal forces that the car is accelerating, which will unbalance the weight on the front and back tires, but the sum of all 4 should be the same as at rest, I would think.

Hmm, need to think more about that...
 
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