Newton's Law of Motion.?

Question

A 8000kg engin pulls a train of 5 wagons each of 2000kg along a horizontal track.If engin exerts a force of 40000N and the track offers a friction force of 5000N.
Then caculate:
(i) force of wagon 1 on wagon 2
(ii)force of wagon 2 on wagon 3
(iii)force of wagon 4 on wagon 5

(i) What is net accelerating froce?
(ii)The acceleration of the train?

Answer 1

An object in motion tends to stay in motion.

Answer 2

NOTE: This answer has been changed.

The engine exerts a force of 40000N that accelerates the whole train. The total force less friction (net accelerating force) is 35000N. This must accelerate a mass of 8000+5*2000= 18000kg The acceleration of the whole train is then 1.944m/sec^2. Every wagon has this acceleration. The friction drag only acts on the wagons, since the force applied by the engine is against the track friction itself.

Consider wagon 5 only. It has a mass of 2000kg, and must accelerate at a under the force it gets from wagon 4, but wagon 4 must also overcome the friction of 1000N. Therefore

f(4,5) = 1.944*2000kg + 1000N

The force on the wagon 4 must accelerate both 4 and 5, so it is just twice f(4,5). So in summary

f(4,5) = 1.944*2000kg + 1000N = 4889N
f(3,4) = 2*f(4,5) = 9778N
f(2,3) = 3*f(4,5) = 14667N
f(1,3)= 4*f(4,5) = 19556N
f(e,1) = 5*f(4,5) = 24444N

EDITED 9/20 TOTALLY REVISED!
EDITED 9/22 Noted net accelerating force.

Answer 3

i) 28000 N + friction force on 4 wagons = 32000 N
ii) 21000 N + friction force on 3 wagons = 24000 N
iii) 7000 N + friction force on 1 wagon = 8000 N

Explanation - since mass of each wagon is the same,
force required to keep each wagon in motion is the same.
there are 5 wagons.
So force extered by engine - friction = force exerted on all 5 wagons = 40000 - 5000 = 35000 N to keep all the 5 wagons in motion excluding effect of friction.

Also since mass of each wagon is the same, friction force on
each wagon is same = 1000 N per wagon

so force exerted on 4 wagons (2,3,4,5) = 28000 N (7000 * 40) +
friction force on 4 wagons = 28000 + 4000 = 32000 N

force exerted on 3 wagons (3,4,5) = 21000 N (7000 * 3) +
friction force on 3 wagons = 21000 + 3000 = 24000 N

force exerted on 1 wagon (5) = 7000 N +
friction force on 1 wagon = 7000 + 1000 N = 8000 N

Hence the answers :

i) 32000 N
ii) 24000 N
iii) 8000 N

Answer 4

First of all we need to distribute the frictional forces. Engine weighs 8000kg and each wagon weighs 2000k. Friction on engine strictly has to be four times the friction on wagon. However for ease of calculation let us assume friction on each wagon tobe F and friction on engine to be 5F
totla friction=10F=5000N given So F=500N
Total engine pull=40000 N
pull available on wagon one=W1=40000-2500(eng.friction)=37500N
Pulll available on wagon two =W2=37500-500=37000N
similarly
W3=36500,W4=36000,W5=35500 All in Newtons

Answer 5

Guessing that you mean a frictional force of 5000 newtons per wagon, it's just 5000 N less per wagon.


Doug

Answer 6

Visualize the problem first. Draw a picture, that helps. Then think what answer would be. Just think about it first for awhile. Not that hard a question. But must think. It is excellent thought question.

Answer 7

If you are kidding OK. If it is requirement you must know it if you passed standard 8.Better you put your calculation for evolution to get satisfied.