01 Mar '17 03:293 edits

I'm going to frame this question in context of an example.

A hoverboard and rider have a combined mass = m

It carries fuel mass = m_fuel

The relative velocity of the exhaust gases are constant = v_exh

A question might be posed: how long can it remain hovering ( a= 0,v=0)

Starting with Newtons 2nd Law: Σ

↑+ Σ

RHS:

m * d

v * dm/dt = 0 ( body is stationary, v=0)

So we have:

-m * g - m_fuel * g

(m+m_fuel) * g = F_thrust ( rearrange and dividng through by -1)

Here is where I'm looking for some clarification. The magnitude of the thrust force is given by: d/dt(m_fuel)*v_exh. The direction ( or sign ) seems to be opposite for Eq1 to be consistent. F_thrust =

Rewriting Eq1:

↑+ Σ

So why the negative? Some books say because the mass flow rate of fuel is decreasing d/dt(m_fuel), I can buy that, but I don't recall ever needing to flip signs in a typical force balance, or that is necessary to define (in advance) the direction of the rate? They typically define themselves in the act of performing the mechanics so long as a consistent coordinate system is used, correct?

What I'm personally gravitating to here is that "F_thrust" is actually a reaction force.

F_thrust + d/dt (m_fuel)*v = 0

Thus, when it is put in terms of d/dt(m_fuel)*v_exh the opposite direction must be assumed.

F_thrust = -d/dt (m_fuel)*v

So, is this a propper way to interpret this?

Performing a change of variables, separating, and integrating finishes out the example, nothing overly exciting there.

Thanks for any help

A hoverboard and rider have a combined mass = m

It carries fuel mass = m_fuel

The relative velocity of the exhaust gases are constant = v_exh

A question might be posed: how long can it remain hovering ( a= 0,v=0)

Starting with Newtons 2nd Law: Σ

**F**= d/dt ( m ***v**)↑+ Σ

**F**= -m * g - m_fuel * g + F_thrust = m * d**v**/dt + v * dm/dtRHS:

m * d

**v**/dt = 0 ( body is not accelerating, a= 0 )v * dm/dt = 0 ( body is stationary, v=0)

So we have:

-m * g - m_fuel * g

**+**F_thrust = 0 Eq1(m+m_fuel) * g = F_thrust ( rearrange and dividng through by -1)

Here is where I'm looking for some clarification. The magnitude of the thrust force is given by: d/dt(m_fuel)*v_exh. The direction ( or sign ) seems to be opposite for Eq1 to be consistent. F_thrust =

**-**d/dt(m_fuel)*v_exh.Rewriting Eq1:

↑+ Σ

**F**= -m * g - m_fuel * g**-**d/dt(m_fuel)*v_exh = 0So why the negative? Some books say because the mass flow rate of fuel is decreasing d/dt(m_fuel), I can buy that, but I don't recall ever needing to flip signs in a typical force balance, or that is necessary to define (in advance) the direction of the rate? They typically define themselves in the act of performing the mechanics so long as a consistent coordinate system is used, correct?

What I'm personally gravitating to here is that "F_thrust" is actually a reaction force.

F_thrust + d/dt (m_fuel)*v = 0

Thus, when it is put in terms of d/dt(m_fuel)*v_exh the opposite direction must be assumed.

F_thrust = -d/dt (m_fuel)*v

So, is this a propper way to interpret this?

Performing a change of variables, separating, and integrating finishes out the example, nothing overly exciting there.

Thanks for any help