This is a great question that brings together a lot of what we ’ve talk about in otherHowStuffWorksarticles about puff . And , it turns out , there is a comparatively simple way to instruct how much drag is on your car .
In the articleHow Force , Power , Torque and Energy Workyou learned about Newton ’s 2d police , which we can restate as force out ( F ) equals mass ( m ) multiplied by quickening ( a ) .
F = ma or a = F/m
atomic number 9 = maora = F / m
What this equation means is that the violence applied to the car will cause your railway car to speed up . When you are ram along at a constant focal ratio , the power produced in theengineis convince to violence at thetires . The drag military unit act in the opposite instruction and is equal to the force that the engine make at the tires . Since these military unit are equal and diametrical , thenetforce on the car is zero , so the car maintains its changeless speed . If you take aside the force produced by the locomotive ( by put the railcar in indifferent , for instance ) then the only force on the car is the drag . Since there is a nett personnel on the car , the car will begin to slow down .
If you could measure the mass of the machine and the acceleration , then you could determine the force . you could have the railway car weighed at alandfillto specify the pot . And you may determine the acceleration by measuring how long it takes the car to slow down when you put it in neutral .
It will help you to understand a little bit about the forces on the car before you define up the experiment .
Theforceto force a gondola down the route deviate with the speed the car is traveling . It follows an equation of the following form :
road load force = a + bv + cv2
The lettervrepresents the velocity of the car , and the lettersa , bandcrepresent three unlike constants :
The important thing about this equation is that the force out on the car begin larger very quickly at high speed . The force on the elevator car at 70 miles per hour is much high than the force at 60 mph .
This means that we require to measure the speedup in a very narrow focal ratio range . Something like 3 miles per hour or 5 km/h should work . We ’ll do this calculation in metrical units because they are easier to work with .
Let ’s say your railroad car has a mass of 2,000 kilograms ( kg ) admit you and your female parent , and you are going to check the acceleration between 100 kilometers per hour and 95 kilometers per hour ( which gives an middling speed of 97.5 kph or about 60 miles per hour so do it on the freeway where the upper limit is high enough ) . You should choose a flat section of route with little traffic , and do it on a Clarence Day when the jazz is serene and it is n’t rain .
Have your mother get the car up to 105 kilometres per hour , and have your stopwatch quick . Tell your female parent to put the car in neutral so you start coast . When the railcar slows down to 100 kph , start the timer , and kibosh it when the car gets to 95 kilometres per hour . You may want to do this several times , peradventure going a different direction on the same part of the freeway . Record all the time and average out them .
allow ’s say the ordinary time was 10 moment . Now you have all the entropy you take to calculate the pull force . You just need to do a few conversions . You need your acceleration in m per minute , per second ( m / s2 ) .
Your car slowed down 5 km/h , which is 5,000 meters per minute , or 1.389 meter per irregular . It took 10 endorsement to slow down this much , so the rate of acceleration was 0.1389 m / s2 .
You just plug away the people and the speedup into the equationF = mato get the force out . There is a ready to hand measuring converterhere .
So the force on this hypothetical railcar at 60 miles per hour is approximately 60 pounds . This also mean that to make the car go 60 miles per hour the railway locomotive has to produce enough business leader to make 60 Sudanese pound of force at the wheels .
We can also figure out how much major power this is . Poweris adequate to force multiplied by speed . So all we have to do is procreate the force in newtons by the speed in meters per second , this gives us the power in watts .
The average fastness of your test run was 97.5 kph , which is 27 cadence per second . So your power is 278 N multiply by 27 m / s = 7,500 watts , or 7.5 kilowatt , which is 10 horsepower .
