A farm tractor tows a 4000-kg trailer up a 19 incline with a steady speed of 3.0

Question

A farm tractor tows a 4000-kg trailer up a 19 incline with a steady speed of 3.0 m/s.

Part A

What force does the tractor exert on the trailer? (Ignore friction.)

Assume that positive x axis is directed toward the direction of motion. Express your answer using two significant figures.

A farm tractor tows a 4000-kg trailer up a 19 incline with a steady speed of 3.0 m/s.

Part A

What force does the tractor exert on the trailer? (Ignore friction.)

Assume that positive x axis is directed toward the direction of motion. Express your answer using two significant figures.

Explanation / Answer

A farm tractor tows a 4000-kg trailer up a 19degree incline with a steady speed of 3.0 m/s .

(No friction mentioned)

Force parallel = mg sin
Force parallel = 4000 kg * 9.8 m/s^2 * sin 19º

Force parallel = 12,762.27 N

Since the farm tractor tows a 4000-kg trailer up a 19degree incline with a STEADY SPEED of 3.0 m/s, ACCELERATION = 0!

Forces = mass * acceleration
ACCELERATION = 0

Forces = 0
Forces = Force of tractor pulling - Force parallel

Force of tractor pulling - Force parallel = 0
Force of tractor pulling = 12,762.27N

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