What is the magnitude of the force?

  • Thread starter Thread starter nmi13
  • Start date Start date
  • Tags Tags
    Force Magnitude
AI Thread Summary
To determine the magnitude of the force in the coupling between the locomotive and the first rail car, start by calculating the total mass being accelerated, which includes the locomotive and both rail cars, totaling 3.00 x 10^4 kg. Using the formula F = ma, where the acceleration is 0.045 m/s^2, the total force exerted by Mike is 1350 N. The force in the coupling can be found by considering the forces acting on the first rail car, which must also accelerate at the same rate. The net force on the first rail car is equal to its mass multiplied by the acceleration, resulting in a coupling force of 675 N. This analysis shows how to approach the problem using free body diagrams and Newton's second law.
nmi13
Messages
4
Reaction score
0

Homework Statement


Mike is training for a strongman competition. Using a rope, Mike is pulling a locomotive of mass 2.00 x 10^4 kg coupled to two 0.50 x10^4 kg rail cars. Mike causes the train to accelerate at 0.045 m/s^2 . What is the magnitude of the force in the coupling between the locomotive and first car?[/B]
Mass of locomotive = 2.00*10^4 kg
Mass of each rail cars = 0.5*10^4 kg
acceleration = 0.045m/s^2

Homework Equations


F= ma

The Attempt at a Solution


I just don't know exactly how to begin to answer this question. I am unable to find any similar questions to help me out with. A nudge in the right direction is all I need. Thank you!
 
Physics news on Phys.org
nmi13 said:

Homework Statement


Mike is training for a strongman competition. Using a rope, Mike is pulling a locomotive of mass 2.00 x 10^4 kg coupled to two 0.50 x10^4 kg rail cars. Mike causes the train to accelerate at 0.045 m/s^2 . What is the magnitude of the force in the coupling between the locomotive and first car?[/B]
Mass of locomotive = 2.00*10^4 kg
Mass of each rail cars = 0.5*10^4 kg
acceleration = 0.045m/s^2

Homework Equations


F= ma

The Attempt at a Solution


I just don't know exactly how to begin to answer this question. I am unable to find any similar questions to help me out with. A nudge in the right direction is all I need. Thank you!

Draw a free body diagram for the rail cars. I.e for the last car, there is a net force pulling the car forward such that the acceleration of the last car is 0.045 m/s^2. What is that force? That force is equal and opposite the force at the back of the first car, pulling back. At the front of the first car, there is also a force pulling it forward. The net force on the first car is such that the first car is accelerating at 0.045 m/s^2. What is the force at the front of the first car?
 
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