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Variable mass beans falling on a platform

  • #1

Homework Statement


A spigot pours beans onto a scale platform.
At a time t = 0.0 sec, the spigot is opened and beans begins to pour out (with initial velocity 0) at a rate of 1.00 kg/sec onto the platform from a height of 10.0 m above.
(a) At t = 10.0 sec, what is the weight of beans on the platform?
(b) What does the scale read at this time?

Homework Equations


It takes ##t=\sqrt{\frac{2h}{g}}\approx 1.43 secs## for each bean to reach the platform.

The Attempt at a Solution


In 10 secs, 10kg of beans will have left the spigot but only ##10/1.43\approx 7kg## have reached the platform. That is the weight of beans in the platform. My understanding is that the scale will read ##7kg##?
 

Answers and Replies

  • #2
kuruman
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10 kg have left the spigot, that's true. If it takes 1.43 s for a bean to drop, how many kg worth of beans are in the air at any particular time?
Also, the scale will read more than the weight of beans that have piled up on it. Why?
 
  • #3
10 kg have left the spigot, that's true. If it takes 1.43 s for a bean to drop, how many kg worth of beans are in the air at any particular time?
Also, the scale will read more than the weight of beans that have piled up on it. Why?
We dont care how many are in the air only how many have reached the platform at t=10s, i.e 7kg?
Good point with the scale reading. I imagine that in addition to the beans that have landed, the impulse of those that landed must contribute although not sure how?
 
  • #4
kuruman
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We dont care how many are in the air only how many have reached the platform at t=10s, i.e 7kg?
The number of beans on the platform is the number that have left the spigot minus the number that is in the air.
You are correct about the impulse. How would you find the force associated with the beans striking the platform? What expressions do you know?
 
  • #5
The number of beans on the platform is the number that have left the spigot minus the number that is in the air.
You are correct about the impulse. How would you find the force associated with the beans striking the platform? What expressions do you know?
Ok the last beans that can reach the platform by ##t=10s## must have left by ##t=8.57s## (if they take ##1.43## to get there) so at ##t=10s##, the mass of beans on the platform is ##8.57kg##.
I know that the velocity of those beans when they hit the platform is ##v=\sqrt{2gh}=14m/s##. So the impulse of the ##8.57kg## will be ##14\times 8.57\approx 120 m/skg##. Is the scale weight ##8.57*9.8+120/10##?
 
  • #6
kuruman
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Not quite right. Sir Isaac Newton said $$F=\frac{dp}{dt}=\frac{d(mv)}{dt}=v \frac{dm}{dt}+m\frac{dv}{dt}$$ Does this give you an idea about finding the impulse force?
 
  • #7
Not quite right. Sir Isaac Newton said $$F=\frac{dp}{dt}=\frac{d(mv)}{dt}=v \frac{dm}{dt}+m\frac{dv}{dt}$$ Does this give you an idea about finding the impulse force?
For part (a) the weight of beans on the platform will be ##8.57\times 9.8\approx 84N##
For part (b) the impulse will be ##F=14\times 1+8.57\times g=14+84=98N##. So the scale will read ##98/9.8=10kg##.
 
  • #8
kuruman
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That is correct. The weight read by the scale is equal to the total weight that left the spigot. Now, do you think this would always be the case or is it a coincidence because of the simple numbers given? You have 10 s for the time, 10 m for the height and 1 kg/s for the rate. Would the reading be something different from "the total weight that left the spigot" if the time were 6.38 s, the height 7.32 m and the rate 0.65 kg/s?
 
  • #9
That is correct. The weight read by the scale is equal to the total weight that left the spigot. Now, do you think this would always be the case or is it a coincidence because of the simple numbers given? You have 10 s for the time, 10 m for the height and 1 kg/s for the rate. Would the reading be something different from "the total weight that left the spigot" if the time were 6.38 s, the height 7.32 m and the rate 0.65 kg/s?
Yes the scale will always read the total weight that left the spigot. Here it is a general argument. We know that ##v=\sqrt{2gh}##, that the mass that has reached the scale by time ##t## is ##m=dm/dt(t-\sqrt{2h/g})##, that ##dv/dt=g## and that the mass that has left the spigot by time ##t## is ##tdm/dt##. So

##F=m\frac{dv}{dt}+v\frac{dm}{dt}=\frac{dm}{dt}(t-\sqrt{2h/g})g+\sqrt{2gh}\frac{dm}{dt}=tg\frac{dm}{dt}## which is the mass that left the spigot during time ##t##.
 
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  • #10
kuruman
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Well done. :smile: Substituting numbers at the very end gives you insights that would be hidden if you substitute up front.
Also note that you dropped a ##g## in your expression. The force should be ##F=gt\frac{dm}{dt}##. It's the "reading on the scale in kg" that is ##t\frac{dm}{dt}.##
 
  • #11
Well done. :smile: Substituting numbers at the very end gives you insights that would be hidden if you substitute up front.
Also note that you dropped a ##g## in your expression. The force should be ##F=gt\frac{dm}{dt}##. It's the "reading on the scale in kg" that is ##t\frac{dm}{dt}.##
Thank you. Fixed the missing ##g##
 

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