Question

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1320 kg/s with a speed of 4.6 104 m/s (at the moment of takeoff).

Answer 1

F=mdv/dt=1320kg/s•4.6104m/s=6085.728kg/ms^2=6085.728N

Answer 2

F=mdv/dt=1320kg/s•4.6104m/s=6085.728kg/ms^2=6085.728N

Answer 3

no, that's not it :/ thanks for trying to help though. I misstyped part of it. instead of 4.6104 it's actually 4.6x10^4. I tried your formula with this number but it still wasn't correct

Answer 4

you can't calculate the net force acting on the rocket unless you know the rocket's mass...

however, the thrust force is given by \[F = v_r \frac{dm}{dt}\]where \(v_r\) is the velocity relative to the rocket and \(\frac{dm}{dt}\) is the rate of mass ejection. When I do it I get a number around 60 million N.

Answer 5

Yes I have also mistyped because right formula. Is F=dm/dtV

Answer 6

Yes I have also mistyped because right formula. Is F=dm/dtV

Answer 7

Mass is variable here

Answer 8

Mass is variable here

Answer 9

problem says at moment of liftoff, so variable mass is not an issue

Answer 10

1320•9.8=12936N

Answer 11

I talked to my instructor and he showed me how to do it. Force is measured in Newtons, which is the same as kgm/s^2, so all I had to do was multiply the two quantities... Guess we all overthought it lol

Answer 12

How

Answer 13

How

Answer 14

@Madison2017 Not sure who you thought "overthought it" — if anything, your instructor has underthought it, as he/she has neglected gravity, which is most certainly exerting force on the rocket at the instant of liftoff. If the problem is restated to ask only for the force exerted on the rocket by the rocket exhaust, then the answer is 60.72 * 10^6 Newtons, as my formula provides.

Answer 15

lol okay. He's not the best insrtuctor, but half of the problem is that the course (AP Physics A/B) is "inquiry based learning". in other words, complete BS

Answer 16

@amity instead of working through equations, all we had to do was look at the units and realize that by multiplying them together, we create the units that make up a Newton, which is used to measure forces.

Answer 17

this is true, but it is not sufficient...what if the correct formula involved a dimensionless constant?

for example, the displacement due to constant acceleration is given by \[x=\frac{1}{2}at^2\]
if you look at the units of \(a\) as [length]/[time]^2 you might jump to the conclusion that you only had to multiply the acceleration by time squared and you would be off by a factor of 2...