Asked by chuck
A 5000kg truck is being pulled up a 15 degree ramp by a cable which is parallel to the surface of the ramp. If the coefficient of friction u between the wheels of the truck and the ramp is 0.100 and the truck is being pulled at a constant velocity, calculate the tension in the cable
Answers
Answered by
Henry
W = m*g = 5000kg * 9.8N/kg = 49,000 N.=
Weight of truck.
Fp = 49000*sin15 = 12,682 N. = Force
parallel to ramp.
Fv = 4900*cos15 = 47,330 N. = Force
perpendicular to the ramp.
Fk = u*Fv = 0.10 * 47,330 = 4733 N. =
Force of kinetic friction.
T-Fp-Fk = m*a
T-12682-4733 = m*0 = 0
T-17,415 = 0
T = 17,415 N. = Tension in rope.
Weight of truck.
Fp = 49000*sin15 = 12,682 N. = Force
parallel to ramp.
Fv = 4900*cos15 = 47,330 N. = Force
perpendicular to the ramp.
Fk = u*Fv = 0.10 * 47,330 = 4733 N. =
Force of kinetic friction.
T-Fp-Fk = m*a
T-12682-4733 = m*0 = 0
T-17,415 = 0
T = 17,415 N. = Tension in rope.
Answered by
Henry
Correction:
Fv = 49,000*cos15 = 47,330 N,
Fv = 49,000*cos15 = 47,330 N,
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