Question 1.
A 6500-kg helicopter accelerates upward at 0.60 m/s² while lifting a 1200-kg car.
(a) What is the lift force exerted by the air on the rotors?
(b) What is the tension in the cable (ignore its mass) that connects the car to the helicopter?
Answer
a)
The total mass that is accelerating at .60 m/s/s is the helicopter mass (6500 kg) and the car’s mass of 1200 kg or 7700 kg.
The total weight would be: (7700 kg)(9.80 N/kg) = 75460 N down (-) and an unknown force by the air on the rotors (F) with an upward (+) acceleration of .60 m/s/s. F = ma looks like:
<F – 75460 N> = (7700 kg)(.60 m/s/s), F = 80080 N upwards = 8.0 x 104 N up
b)
The tension in the cable is affected by the car’s mass, so looking at the 1200 kg car we have this situation:
The total weight would be: (1200 kg)(9.80 N/kg) = 11760 N down (-) the tension in the cable upward (T) with an upward (+) acceleration of .60 m/s/s. F = ma looks like:
<T – 11760 N> = (1200 kg)(.60 m/s/s), F = 12480 N upwards = 1.2 x 104 N up
Question 2.
A city planner is working on theredesign of a hilly portion of a city. An important considerationis how steep the roads can be so that even low-powered cars can getup the hills without slowing down. It is given that a particularsmall car, with a mass of 1100kg, can accelerate on a levelroad from rest to 21 m/s (75.6km/h) in13.0s. Using this data, calculate themaximum steepness of a hill.
Answer
max acceleration =v/t=(21m/s)/(13s)=1.61538 m/s/s
the max force the car can apply is ma=(1100kg)(1.61538m/s/s)=1776.923N
Now with the hill, we know the downward force will be mass X gravity which is (9.8 m/s/s)(1100kg)=10780 N
so the car moves up the incline at 1776 N and gravity pulls down with 10780 N.
the sine of an angle is opposite/hypotonuese
Sin θ=(1776.9N)/(10780N)=.165588
θ=9.5314 degrees
