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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

KingMongolian
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1 Answers1

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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.

Brain Stroke Patient
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