A tractor/trailer truck on a 7% downgrade (road slanted at 4 degrees below the horizontal ) is limiting its acceleration to 0.3m/s per second by braking. If there were no breaks on the 10000kg trailer, how large a backward force would the tractor be applying to the trailer in this situation?
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Acceleration on 4 deg. slope for trailer without brakes = (sin 4 deg x 9.8), = 0.684m/sec^2.
Difference is acceleration by braking with tractor = (0.684 - 0.3) = 0.384m/sec^2.
Force being applied by tractor = (mass x acceleration) = 10,000 x 0.384, = 3,840N.
Brakes aren't "breaks". The trailer might break if overloaded.?
Difference is acceleration by braking with tractor = (0.684 - 0.3) = 0.384m/sec^2.
Force being applied by tractor = (mass x acceleration) = 10,000 x 0.384, = 3,840N.
Brakes aren't "breaks". The trailer might break if overloaded.?
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lol right the word is brakes
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what about this on :
A 150kg trailer is pulled with a horizontal force of 550N and accelerates at 3m/s^2. How large is the frictional force tending to hold the trailer back?
please do me a favor by solve it and i will be very appriciate to u :)
A 150kg trailer is pulled with a horizontal force of 550N and accelerates at 3m/s^2. How large is the frictional force tending to hold the trailer back?
please do me a favor by solve it and i will be very appriciate to u :)
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