can u please solve this for me?

Question

A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg,
along a horizontal track. If the engine exerts a force of 40000 N
and the track offers a friction force of 5000 N, then calculate the force of wagon 1 on wagon 2.

Answer 1

28 000 Newtons. Common give me the ten points! The engine provides a net force of 35 000N. 7000N is used by the first wagon and you are left with 28000N. I must warn you that there are too forms of friction. The first one is static friction stopping the bogy from moving. Once the body starts moving then you will be dealing with dynamic friction which is a lot less than its counterpart.
In this case the force on the second wagon will turn up much greater than 28000N. It may be about 30000N when the body starts moving.
Friction is not a function of surface area of contact. So we can assume it is acting on the engine alone!

Answer 2

No matter how large the tractive force may be, the engine moves by the purchase it can gain on the track due to friction between the wheels and the track. So the maximum force, with which the engine can pull the wagons, in this instance, is 5000 N. The excess force (torque) causes the wheels to turn and slip on the track.

The wagons will roll or slide on the track depending on the magnitude of the pull of the engine and the frictional force between the wheels and the track. Assuming proportionality of the frictional force and the weight of the wagons, the maximum frictional force of each wagon and the track is (2000/8000)*5000 = 5000/4 N. The 1st wagon which is pulled by the engine with a force of 5000 N will therefore roll and also slide on the track.
The pull of the 1st wagon on the 2nd wagon will be 5000 N minus the sliding friction between the wheels of the 1st wagon and the track.

Answer 3

this should be sorted out in this way:

mass of the total train=2000*5+8000=18000
net force working on the train=40000-5000=35000
[here i am considering the friction as the friction over the entire body]
we know,
ΣF=(Σm)a
=>a=ΣF/Σm=35000/18000=1.944

as all thw wagons are directly attached with the engine, they will have the same acceleration, so accleration on the wagon 2 will be 1.944

so ΣF=ma=2000*1.944=3888.88 N

as considered in the question we can show two forces in opposite direction are workinf on the wagon,
1) friction force(F')
2) force from the first wagon(F")

so F"-F'=3888.88
=>F"=3888.88+F'

now friction on a body is proportional to its mass, if the frictional force on the total mass equates 5000 N, then wagon 2 will encounter a force of 5000*2000/18000=555.56 N

so the force created by wagon 1 on wagon 2 will be
F"=3888.88+555.56=4444.44 N

i think it should be right.

Answer 4

Interesting, I see the answer as quite different from those already given. Given that forces are transmitted along a rigid body, like a train, I see the solution as:

<==40000N - 5000X6N==> acting on a total mass 8000kg + 5X2000kg (There are six units on the train, each with rolling friction force.)

Thus a = 10000/18000 = 5/9 m/sec-sec for the entire train.

So the net (forward) force on any wagon is F = ma = 2000X5/9 = 10000/9 ~ 1111N and for the locomotive F = Ma = 8000X5/9 ~ 4444N

As a check, sum up the net forces forward on each unit. 5555N for the five wagons, plus 4444N for the locomotive = 9999 ~ 10000N for the entire train. This is consistent with the 40000N motive force minus the sum of six 5000N rolling friction forces operating in opposition.

PS: I should explain my assumption that the locomotive also had 5,000N of frictional force. 5,000N would be true if and only if its coefficient of rolling friction was 1/4 that of each wagon. Why? Because F(rolling) = k(rolling)(mg); since the mass of the locomotive is four times the mass of each wagon, its rolling friction force could only be the same as each wagon's when its k value is 1/4 of each wagon's k value.

Answer 5

x² = x+6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 So x = 3 or - 2 x² + 2x = 0 x(x + 2) = 0 So x = 0 or - 2 60 = x² - 4x x²- 4x - 60 = 0 (x - 10)(x + 6) = 0 So x = 10 or -6 4x = x² - 40 5 x² - 4x - 40 5 = 0 (x - 9)(x + 5) = 0 So x = 9 or -5 -6x = x² + 9 x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 (x + 3)² = 0 So x = -3 -15=x² + 8x x² + 8x +15 = 0 (x + 5)(x + 3) = 0 So x = -5 or -3 -9x = x² + 14 x² + 9x + 14 = 0 (x + 7)(x + 2) = 0 So x = -7 or - 2 wish that helps. =D

Answer 6

mass of engine =8000kg
mass of 5 wagons= 2000 of 5= 10000 kg
Force exheted by engine= 40000 N
frictional force = 5000 N
net force = force exherted by engine-- frictional force
= 40000-- 5000 N
= 35000 N
and net acceleration = net force/ mass of engines
= 35000/10000
= 3.5
now,
force of wagon 1 on wagon 2 = 35000-- 3.5 of 2000(which is mass of 1 wagon)
= 35000-- 7000
= 28000 N

Answer 7

|---------------------| |---------------------|
| wagon 1 |-- | wagon 2 |--
|-------------------- | |---------------------|
N12 <------ N23 <------

net accelaration = 40000-5000/18000=35/18 N
let the contact foece offered by wagon 4 on wagon 5 b N45 and similarly
N45=2000*35/18
similarly
N12=N23+ma=N34+2ma=N45+3ma=4ma
=4*N45=8000*35/18=15555.55N

Answer 8

sorry buddy, i forgot how to do that. What grade are you in? I think i took this my sophmore year in chem. Phys.

Answer 9

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Answer 10

Is this one of your examination questions?