# Different forces from different inertial frames?

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? • Yes the velocity is constant and I don't really see how what you wrote above resolves the issue. – Brain Stroke Patient Jul 19 at 10:32
• 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 ? – Eli Jul 19 at 10:40
• Edited question with link – Brain Stroke Patient Jul 19 at 10:49

• $F=\frac{dp}{dt}$ 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. – gandalf61 Jul 19 at 13:03
• 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 $m$. – gandalf61 Jul 19 at 15:39