Solving the Force of Reverse Thrust on Mars' Polar Lander

[Nx2]

Hi guys… there was a lot of arguing in my class when my teacher gave us this one assignment question that I’m not too sure about. It goes as follows…

A Polar Lander somewhere above the surface of the Mars is landing with a deceleration of 23m/s^2. If the Lander itself had a mass of approximately 700kg, solve for the net force.

So, this is what I did:
Fnet = ma
= (700kg)(-23m/s^2)
= -16100N [down] or 16100N [up]

Ok so this is the weird part, there was a second part to the question which goes as follows…

The acceleration due to gravity of Mars is 3.6m/s^2. Using this fact solve for the force of reverse thrust that was required to provide the necessary overall upward net force found above.

… there was a diagram attached as well which helped make it easier but basically what is showed was an arrow downwards on the Lander saying it was Fg and an arrow opposite to Fg (upwards) which was the force of the reverse thrust slowing it down, named Fr.

So this is what I did:

Fg = mg
= (700kg)(3.6m/s^2)
= 2520N

And this is where all the arguing started. We weren’t too sure what formula to use to find Fr.

Fnet = Fg – Fr or is it Fnet = Fr- Fg? He said the answer is 18 620 N

This is what I did. Using Fnet = Fg - Fr:
Fr = -Fnet + Fg
= -(-16100N) + 2520N
= 18 620 N which looks correct to me.

But when I saw the answer done for us, this was the method used:

Fnet = Fr – Fg
Fr = Fnet + Fg
= (16100) + 2520 (NOTE: notice how Fnet all of a sudden became positive?)
= 18 620 N
Can you do that?... or is my way the correct way? I just don’t want to lose marks for something that was given to us but seemed wrong through my eyes.

 

[jamesrc]

You need to start off with your free body diagram and let that be the way you determine what signs go where. For example, I would draw the lander as a block with Fr pointing up and Fg = mg pointing down. In my coordinate system, I am specifying that up is positive. Since the lander is decelerating, it is accelerating in an upward direction. This means that the net force points upward as well. So now you can write Fnet = Fr - Fg where Fnet = ma and a is a positive number (the magnitude of the deceleration). I don't think I wrote that very clearly; let me know if it still doesn't make sense.

 

[Nx2]

yea i get what u are saying. so if up is positive in the diagram then the Fnet since it is slowing down(moving up in this case) will also change to positive?... so if i left it the way it was where Fg was positive would this wouldn't be wrong would it?

Fnet = Fg - Fr
Fr = -Fnet + Fg
= -(-16100N) + 2520N
= 18 620 N

 

[jamesrc]

It's not "wrong" provided that you understand that you have now defined down as the positive direction.

 

[Nx2]

alright kool... thnx alot

- Tu