01 Mar '17 03:29>3 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: ΣF = d/dt ( m * v )
↑+ ΣF = -m * g - m_fuel * g + F_thrust = m * dv/dt + v * dm/dt
RHS:
m * dv/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:
↑+ ΣF = -m * g - m_fuel * g - d/dt(m_fuel)*v_exh = 0
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: ΣF = d/dt ( m * v )
↑+ ΣF = -m * g - m_fuel * g + F_thrust = m * dv/dt + v * dm/dt
RHS:
m * dv/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:
↑+ ΣF = -m * g - m_fuel * g - d/dt(m_fuel)*v_exh = 0
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