Bubba is trying to pull his boat out of the lake after a rainy day fishing, but

Question

Bubba is trying to pull his boat out of the lake after a rainy day fishing, but the rain has made the boat ramp muddy and slick enough that he can't back his truck down the hill to attach the boat trailer. Instead he attaches a rope to the back of his truck, runs the rope over a parking barrier, and attaches the other end to the trailer. He can then pull the boat and trailer up the hill with the truck on level ground.

Unfortunately, the rope he uses to do this is the rope that usually attaches the boat to the trailer. The boat is just sitting on the trailer with nothing (but friction) to keep it in place. Of course there is also friction between the trailer and the ramp. Air drag is zero.

What you know is:

Mass of Boat = 140 kg

Mass of Trailer = 380 kg

Mass of Truck = 2000 kg

Angle of ramp = 6 degrees

Coefficent of Static Friction, Boat on Trailer = 0.55

Coefficient of Kinetic (or rolling) Friction, Ramp on Trailer = 0.25

What is the maximum tension that Bubba can give to the rope and get his boat up the ramp without it sliding off the trailer?

Give your answer in Newtons to at least three significant figures

Explanation / Answer

On boat :

perpendicular to the incline, N = 140gcos6

friction, = uN = 0.55 x 140 x 9.8 x cos6 = 750.47 N

friction is the force that will help boat to go along the trailer.

so first we calculate what maximum acceleration it can have.

f - mg = ma

750.47 - (140 x 9.8 x sin6) = 140a

a =4.34 m/s^2

so acceleration of boat trailer system is 4.34 m/s^2 .

On tailer-boat system,

perpendicular to the incline,

N - mgcos6 = 0

N = mg cos6

friction =u N = 0.25 mg cos6 = 0.25 x (140 + 380) x 9.8 x cos6

f = 1267.02 N

along the incline,

T - f - mgsin6 = ma

T - 1267.02 - (140 + 380) x9.8 xsin6 = (140 + 380) x 4.34

T - 1800.24 = 2256.8

T = 4057.04 N . ......Ans

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