Calculating Lift Force of Helicopter Propellers

[chemguy990]

Homework Statement

A 5400 kg helicopter accelerates upward at 2.17 m/s2. The acceleration due to gravity is 9.8 m/s2. What life force is exerted by the air on the propellers? Answer in N.

Homework Equations

Fnet=ma

The Attempt at a Solution

I am making the acceleration positive (2.17m/s2) and the gravitational acceleration negative (-9.8 m/s2) since the helicopter is traveling upwards.

F(lift) - F(gravity) = ma
F(lift) - (5400kgx(-9.8m/s2)) = (5400kg)(2.17 m/s2)
F(lift) = 11718-52920 N
F(lift) = -41202 N

Shouldnt the lift force acting upon the propellers by positive?

 

[arildno]

Sure!
Net force means the SUM of all forces, positive and negative.
Thus, you have a minus too much here:

F(lift)+F(grav)=ma
F(Lift)=ma-F(grav), remember that F(grav) is here a negative quantity.

 

[m2003]

Yes, the lift force should be positive since it is acting in the same direction as the acceleration of the helicopter. Therefore, the correct solution should be:

F(lift) = ma + F(gravity)
F(lift) = (5400 kg)(2.17 m/s^2) + (5400 kg)(9.8 m/s^2)
F(lift) = 11718 N + 52920 N
F(lift) = 64638 N

The lift force exerted by the air on the propellers is 64638 N. This is the force that is responsible for lifting the helicopter off the ground.