I've been having trouble understanding this equation here derived in the picture given below. If I change reference frames to one which is moving at a velocity u with respect to the first frame, then the term $v$ in the equation would change to $(v-u)$ however as far as I can tell the $\frac{dm}{dt}$ can't change. Does this means these two observers would measure different values of force? So far, I've been told that all inertial reference frames measure the same value for a given force.

Also isn't $F=\frac{dp}{dt}$ only valid for systems with mass that is not varying with time? As they said in here: Second law of Newton for variable mass systems

If so, then why is the text considering $\frac{dp}{dt}$ to be the force?

An introduction to Mechanics-Kleppner and Kolenkow

  • $\begingroup$ Yes the velocity is constant and I don't really see how what you wrote above resolves the issue. $\endgroup$ Jul 19, 2019 at 10:32
  • $\begingroup$ sorry I write it again with $P=(M+m(t))\,v$ and $F=dP/dt$ you get $dP/dt=dm/dt\,v$ this is your result ? $\endgroup$
    – Eli
    Jul 19, 2019 at 10:40
  • $\begingroup$ Edited question with link $\endgroup$ Jul 19, 2019 at 10:49

1 Answer 1


Okay I got my answer. If I change reference frames to a frame moving with velocity u with respect to the first frame, then the sand is no longer falling vertically but rather it will have a horizontal velocity component of -u in the second frame. So that the equation of motion would be

F = [(v-u)-(-u)]dm/dt = vdm/dt

Also, F is considered to be equal to dp/dt because they considered the whole sand + truck as the system so that the system isn't really losing any mass overall. Smh i am so slow at understanding things.

  • $\begingroup$ $F=\frac{dp}{dt}$ is always valid in an inertial frame. $F=ma$ is only valid if $m$ does not change with time. You can consider the freight car and the hopper separately, but there is no horizontal force on the hopper because the sand falls vertically w.r.t. the hopper - the horizontal momentum of the sand (in any intertial frame) does not change until it hits the freight car. $\endgroup$
    – gandalf61
    Jul 19, 2019 at 13:03
  • $\begingroup$ If F=dp/dt is always valid then are the answers provided in the link I gave with my question wrong or am I missing something? Btw I'm strictly speaking about Newtonian mechanics, I get how things are different in special relativity. $\endgroup$ Jul 19, 2019 at 15:25
  • $\begingroup$ The linked answer looks fine. In the frame of reference in which the hopper is stationary, a force must be applied to the sand to increase its velocity from zero to the velocity of the freight car. In the frame of reference in which the freight car is stationary, the same force must be applied to the sand to increase its velocity from an initial negative value (the hopper is now going backwards) to zero. In either case the force is $v \frac{dm}{dt}$ which is derived from $\frac{d}{dt}(mv)$ with constant $v$ and changing $m$. $\endgroup$
    – gandalf61
    Jul 19, 2019 at 15:39
  • $\begingroup$ No, i mean the linked answer does say F is not always equal to dp/dt but you said F = dp/dt is always valid in an inertial frame. I need some clarification here, because the two statements seem contradictory to me. Besides if F is always equal to dp/dt then then for both variable mass and velocity F would be mdv/dt + vdm/dt . But that's not the case, right? For example, for rockets this equation isn't true. $\endgroup$ Jul 19, 2019 at 16:38

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