Inclined Plane

[mushroom25]

I have a basic physics problem at work, but since it's 30 years since I did any of this stuff, I'm struggling. Need to tow a 46000kg truck up a 6% gradient & work out power required in watts.

46000x9.81x1.06 = 478,335/0.6 (coeff of friction) = 797226N

Power = forcexdistance/time, or forcexvelocity
797,226x0.5 = 398,613Nm/sec - 1J = 1Nm, = 398.6 kW.

That's getting on for 500hp for one lousy truck - obviously I've made a mistake, but where?

can anyone spot my error?
(thanks)

[Vagrant]

A normal road truck of that weight would be about 500hp - and could probably achieve 30-40 mph on that slope.

I presume the speed you want to ascend at needs taking into account.

[igtretd]

mushroom25..it's about the same length of time since I did any of this but I think there are a couple of errors in your calculation.

Firstly the weight component down the slope is mgsin(angle of slope)

For a 6% slope the sine of the angle of the slope is approximately 0.06, so in rounded numbers the weight component down the slope is in the order of 46000kg x 10 x 0.06 =27000kg which when you factor in the coefficient of friction of 0.6 becomes 46000kg!

Secondly work = force x distance moved. Lets say you want to move the truck 20m then the work is 46000kg x 20m = 0.9MJoules.

A Joule is a watt per second so if you wanted to take 60s to move the truck 20m up the slope you would have to supply 0.92MJoules in 20s or 920,000/20 = 46kW i.e. about a 50HP motor.

[igtretd]

ooops...46kW is 61 HP so allowing a margin for overcoming initial inertia/stiction a 100HP motor to achieve walking pace up the slope