Space Shuttle Lift Off: Acceleration and Time Calculations for Homework

[Kryslynn]

Homework Statement

At lift off, the three main engines and the two booster rockets of the 4.5X106 lb space shuttle produced 6.4X106 lbs of thrust vertically downward. What was the acceleration at liftoff and assuming the acceleration remains constant how long did it take the shuttle to rise through its own height of 184 ft?

Homework Equations

S=So+V*t+1/2*a*t^2
V=Vo+a*t

The Attempt at a Solution

Fdown = 4.5 x 106 lb., F Thrust = 6.4 x 106 lbs.
lb. = slug x ft. /sec2
6.4 x 106 lbs - 4.5 x 106 lb = 1.9 x 106 lbs. upwards
F=mg --> m = f/g.
(1.9 x 106 lb.)/(32 ft./sec2) = 5.8 x 104 Slugs
F=ma --> a = f/m

V = 0 + 32 ft./sec2

 

[Doc Al]

Fdown = 4.5 x 106 lb., F Thrust = 6.4 x 106 lbs.
lb. = slug x ft. /sec2
6.4 x 106 lbs - 4.5 x 106 lb = 1.9 x 106 lbs. upwards

Good. You found the net force.

F=mg --> m = f/g.
(1.9 x 106 lb.)/(32 ft./sec2) = 5.8 x 104 Slugs

Here you made an error. Mass = weight/g, not (net force)/g.

(That's why your calculated acceleration was equal to g.)

 

[Kryslynn]

Thank you so much! That makes sense. =)