Train Moving pulling two boxcars

  • Thread starter Thread starter Jordash
  • Start date Start date
  • Tags Tags
    Train
AI Thread Summary
The discussion revolves around calculating the force exerted by the first boxcar on the last boxcar in a train system. The train consists of a locomotive and two boxcars, with a total mass of 28,000 kg for the boxcars and an acceleration of 0.4 m/s². The initial calculation mistakenly combines the masses of both boxcars, leading to an incorrect force of 11,200 N. The correct approach involves using Free Body Diagrams (FBD) to isolate the last boxcar and apply Newton's second law, resulting in a force of approximately 6,000 N. Understanding the distinction between the forces acting on the entire system versus individual components is crucial for accurate calculations.
Jordash
Messages
64
Reaction score
0

Homework Statement


A train consists of a 4000 kg locomotive pulling two loaded boxcars, the first with a mass of 12,000 kg and the second with a mass of 16,000 kg. Assume that the boxcar wheels roll without friction and ignore aerodynamics. The acceleration of the train is 0.4 m/s2.

With what force strength, in Newtons, does the first boxcar pull on the last boxcar?


Homework Equations



F=ma

The Attempt at a Solution



I thought the mass should be the total of the last two boxcars which would be 12000 plus 16,000 =28,000 so I tried 28,000*0.4m/s^2 which came out to 11,200 which is obviously wrong the answer to the question should be 6E3, what am I doing wrong?
 
Physics news on Phys.org
If the train is accelerating then what force is accelerating just the last car?
 
Last edited:
Jordash said:

Homework Statement


A train consists of a 4000 kg locomotive pulling two loaded boxcars, the first with a mass of 12,000 kg and the second with a mass of 16,000 kg. Assume that the boxcar wheels roll without friction and ignore aerodynamics. The acceleration of the train is 0.4 m/s2.

With what force strength, in Newtons, does the first boxcar pull on the last boxcar?


Homework Equations



F=ma

The Attempt at a Solution



I thought the mass should be the total of the last two boxcars which would be 12000 plus 16,000 =28,000 so I tried 28,000*0.4m/s^2 which came out to 11,200 which is obviously wrong the answer to the question should be 6E3, what am I doing wrong?
You should be using Free Body Diagrams (FBD). What you have calculated is the the pulling force of the engine on the first boxcar, not the pulling force of the first boxcar on the last boxcar. The easiest way to do this problem is to look at the last boxcar only in a FBD of that last boxcar. Identify the force acting on it, and solve for its value using Newton 2.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top