Calculating Thrust Force of a Model Rocket

AI Thread Summary
To calculate the thrust force of a model rocket with a mass of 0.48 kg accelerating at 34 m/s² and facing an air resistance of 2.4 N, the net force must be determined using Newton's second law. The forces involved include the thrust force (Fr), air resistance (Fa), and gravitational force (Fg). The gravitational force is calculated as Fg = -0.48 kg * 9.8 m/s², resulting in Fg = -4.704 N. By applying the equation Fr + Fa + Fg = m*A, the thrust force is found to be Fr = 23.424 N, which can be rounded to 23.4 N in SI units. Properly analyzing the forces and their directions is crucial for accurate calculations.
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1. Homework Statement

A model rocket of mass 4.8 x 10^2 g accelerates vertically upward at 34.m/s2.
Calculate the thrust force applied by the rocket if the air resistance is 2.4N

2. Homework Equations

F=ma?


3. The Attempt at a Solution

Converted into SI
rocket mass = 0.48g

Calculated the Force of the Rocket
F=ma
F= 16.32N

Stuck here can anyone help?
 
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First make up names FOR EACH SEPARATE FORCE YOU CAN THINK OF AS RELEVANT IN THE PROBLEM!
 
Like...
Fr = Rocket Force
Fa = Air resistance
Fg = Gravity
?
 
Indeed!
Now, set up Newton's 2.law of motion where you include all of these forces EXPLICITLY!
 
Im not sure how to find the force of gravity though, is it 9.8?
 
Can you please set up in SYMBOLS Newton's 2.law?
 
Like what do you mean? Like F=ma?
 
xRadio said:
Like what do you mean? Like F=ma?

Written with the symbols you made up for the three separate forces, please.
 
Fr = (0.48kg)(34)
= 16.32 N

Fa = ma
2.4N = ma

Fg = ma
 
  • #10
Totally wrong!

You have to learn that we SUM FORCES TOGETHER TO GET THE NET FORCE F!

Thus, Newton's 2.law reads (using big A for acceleration of rocket):
Fr+Fa+Fg=m*A

Do you understand this?
 
  • #11
yes, ohh, i thought that since Fa and Fg was going the opposite direction that it would be subtracted from Fr.
How would i go from there
like this?

Fr+Fa+Fg=m*A
16.32 + 2.4 + Fg = m(34)

is this correct?
 
Last edited:
  • #12
ALWAYS start with setting up Newton's 2.law with names for the separate forces before you start to simplify the "force side" of it by invoking directions of forces, specific force laws etc.

NOW, let us continue:
If we let the positive direction be away from Earth, what are the signs of Fr,Fa,Fg and A?

I saw your EDIT:
It is incorrect
 
Last edited:
  • #13
Fr = +
Fa = -
Fg = -
A = +
 
  • #14
Correct!
So, Fa=-2.4N with direction included, agreed?
 
  • #15
Yup, agreed.
 
  • #16
Now, the force of gravity is nothing else than the WEIGHT of the rocket.
So, with direction, what is Fg?
 
  • #17
Fg = -0.48N?
 
  • #18
Weight is given by (mass)*(acceleration of gravity), that is with units and direction, we have:
Fg=-0.48*9.8N

Now, insert all known quantities into their proper place in Fr+Fa+Fg=mA!
(Skip the units at this point, if you like)
 
  • #19
Fr = ?
Fa = -2.4 N
Fg = -4.704 N
m = .48 kg
A = 34 m/s2

Fr =.48*34 +2.4 + 4.704
Fr = 23.424 N

Correct?
 
  • #20
Probably, your logic is completely sound, so it is only calculator mistakes you might have made.
 
  • #21
OK thanks alot.
 
  • #22
Notice that your problems readily solved themselves when you first started to split your problem into pieces in a systematic manner. This is called "to analyze a problem".
 
  • #23
Alright thanks a lot for your help arildno.
What would the answer be if it had to be put in SI digits?
Fr = 23.424 N
Fr = 23.4 N ?
 

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