Solving the Two-Car Train Acceleration Problem

AI Thread Summary
The discussion addresses the problem of calculating the tension in a two-car train system under acceleration. It establishes that the tension (T1) between the locomotive and the first car is twice that of the tension (T2) between the first and second cars. This conclusion is derived using Newton's laws and considering the forces acting on the system. By analyzing the forces on both the entire train and the last car, the relationship T1 = 2T2 is confirmed. The explanation includes a diagram suggestion to visualize the forces involved.
Elvis
Messages
14
Reaction score
0
Hi !

can anyone help me ?

A train locomotive is pulling two cars of the same mass behind it. Show that the tension in the coupling between the locomotive and the first car is twice that between the first car and the second car, for any nonzero accelerationv of the train .

thanks
 
Physics news on Phys.org
Draw yourself a little diagram of the situtation. Include the forces acting. You should have a force(say, F) coming out of the loco, between the loco and the first car call the tension T1; remember to draw two arrows for this tension, one pulling the loco towards the car, the other pulling the car towards the loco (these have the same magnitude (By Newton's Third Law) of T1. Similarly, draw in T2 for the tension between the cars. Let the mass of each car be m.

All the bodies (both the cars and the loco) must have the same ACCELERATION, otherwise they would split apart or crash. Call this accleration a.

Consider an imaginary box around the two cars, which bisects the bar holding the first car to the loco. Now find F = ma horizontally, taking to the right as positive:

T1 + T2 - T2 = (2m)a
i.e. T1 = 2ma

Similary, imagine a box around the LAST car, which bisects the bar holding the first car to the last car. We find:

T2 = ma

hence, T1 = 2T2.

Hope this is sufficient for your needs.
 
Thank you Rudipoo .
 
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...

Similar threads

Back
Top