Example 12: A 1000 kg engine pulls a train of 4 wagons each of 2500 kg along a horizontal railway track. If the engine exerts a force of 50000 N on wagons and track offers a force of fiction of 2500 N on each wagon, then calculate (i)The net accelerating force (ii) the acceleration of the train (iii) the force by Wagon 1 on wagon 2 I am unable to understand the solution to this example. Pls, help as you find the solution image below. This question is from Aakash institute's book.
Answers
Answer.
Complete step by step answer:
The FBD of the given system can be drawn as shown in the figure below.
Given that,
Mass of engine me
= 1000 kg
Mass of wagon mw
= 2500 kg each
Force by engine F
= 50000 N
Friction on each wagon f
= 2500 N
Now, we know that the friction force is applied in the opposite direction of acceleration as shown in the figure above. The friction force of 2500 N is applied on each wagon
Therefore, net friction force f
is
fnet=2500×4
Therefore,
fnet=10000N
Therefore, we know from figure that net force on the system will be given by,
Fnet=F−fnet
After substituting the values
We get,
Fnet=50000−10000
Therefore,
Fnet=40000N
………….. (1)
Now, to calculate the acceleration of train
We know from Newton's law that,
F=ma
Here, m is the total mass that is the sum of mass of engine and all the four wagons
Therefore,
m=me+4mw
After substituting the given values
We get,
m=1000+4×2500
On solving,
m=11000kg
…………. (2)
Therefore, from (1) and (2) we can say that
40000=11000×a
Therefore,
a=3.63m/s2
………………… (3)
Now, from the figure we can say that force exerted by the wagon 2, 3 and 4 will be equal to force exerted by wagon 1 on wagon 2.
From (3) we can say that acceleration of all the wagons is 3.63. Now the force exerted by wagon 1 on wagon 2 can be given by,
F12=3×mw×a
……………. (since force by wagon 1 on 2 will equal force by other wagons on 1)
Therefore,
F12=3×2500×3.63
Therefore,
F12=27225N