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A flat car of mass $m_0$ starts moving to the right due to a constant horizontal force $F$. Sand spills on the flat car from a stationary hopper. The rate of the loading is constant and is equal to $\mu$, with units $\mathrm{kg/s}$. What is the acceleration of the car in the process of loading as a function of time?

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I tried applying F=m*a to this problem, but it didn't work: $F/(m_{0}+\mu*t)=a$ is incorrect. I know that there is something called a variable mass system, but I am unable to understand my why answer, intuitively, is incorrect. Any ideas?

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Your intuition wasn't accounting for an important factor, and that is why you got the incorrect answer.

If you're only considering the mass at the cart at any point in time, and the acceleration required to move a cart of that mass, your intuition would be correct. There is more going on than that though, and this is what seems to have been missed by your intuition.

Consider what is happening to the falling sand. The sand is falling without any horizontal velocity, but the car is moving horizontally. For the sand to move with the cart, it needs to gain a horizontal velocity. To analyze how this might change the motion, it's easiest to consider conservation of momentum.

At any given point in time, sand is falling, and that sand requires a change in momentum to move with the car. This means that the faster the car is moving, the greater the impulse that has to be added to the falling sand along with the increased resistance to movement that is induced by the increased mass.

This adds another term to the relationship between acceleration and time, and this term will have to depend on the cars velocity. This leads to a differential equation relating the two.

An example of the derivation can be found on the Wikipedia page for Variable-mass system (accretion).

I will leave the specific derivation of this answer to you.

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  • $\begingroup$ Ah I got you, so basically, some of the force goes to bringing up the sand to speed correct? $\endgroup$ – Dude156 Jul 31 '19 at 18:44
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    $\begingroup$ @Dude156 Exactly. And it's not the same as the force that goes into increasing the speed of the sand+car system, which is all that you calculated. It's a bit of a weird one to think about at first, but it's pretty obvious once you see the full picture. $\endgroup$ – JMac Jul 31 '19 at 18:46
  • $\begingroup$ Yep, that would be the friction force that's speeding up the sand, but slowing down the car right? $\endgroup$ – Dude156 Jul 31 '19 at 18:47
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    $\begingroup$ @Dude156 Friction and collisions I would assume. It would be tough to categorize the forces specifically I would say (all the interactions between falling sand and the cart might get complicated); but you can look at the bulk and just think Newton's third law. The car applies a force on the sand to get it up to speed; the sand has an equal and opposite reaction which slows the car down. $\endgroup$ – JMac Jul 31 '19 at 18:50
  • $\begingroup$ As I tried solving this, I ended up with a differential equation. Any ideas how to proceed from here? $\endgroup$ – Dude156 Jul 31 '19 at 20:05
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The problem here is that Newton's 2nd law is actually $F=\frac{dp}{dt}$ for momentum $p$. In the constant mass case this just gives you $F=ma$ but, if mass is not fixed then we get $F=\mu v+ma$(using the product rule) (with $\mu=\frac{dm}{dt}$ and $a=\frac{dv}{dt}$, which we can can rewrite as $a=\frac{1}{m_0+\mu t}(F-\mu v)$ This is because a certain force $\mu v$ is required to accelerate the new mass up to speed $v$ before further acceleration can be considered.

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  • $\begingroup$ the mass in the second law is a constant and you cannot take the derivative that way $\endgroup$ – Wolphram jonny Jul 31 '19 at 18:43
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    $\begingroup$ The use of $F=dp/dt$ for variable mass systems is more involved than just applying the product rule for derivatives. $\endgroup$ – Aaron Stevens Jul 31 '19 at 19:49

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