Question
Question: A 1000 kg engine pulls a train of 4 wagons of 2500 kg each along a horizontal railway track. If the ...
A 1000 kg engine pulls a train of 4 wagons of 2500 kg each along a horizontal railway track. If the engine exerts the force of 50000 N on wagons and track offers force of friction 2500 N on each wagon, then calculate 1) The net accelerating force. 2) The acceleration of the train. 3) the force of wagon 1 on wagon 2.
Solution
In this question, we have been asked to calculate the net accelerating force, the acceleration of train and the force of wagon 1 on wagon 2. We know that the force is given as mass times acceleration. Therefore, we shall calculate acceleration using Newton’s law. Now, we also know that the friction force opposes motion. Therefore, it will be in the opposite direction of acceleration of the train. Therefore, net force can be given by the difference between the force by engine and the friction force.
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
Note:
From the three laws given by Newton, the second law states that force acting on an object is equal to the product of mass of the object and the acceleration of the same object. Therefore, if the object has a greater mass, greater force will be required to move or accelerate the object. This law can also be stated as the force acting on an object is given as the mass times rate of change of velocity of the object.