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I How much force to throw an object vertically?

  1. Oct 31, 2018 #1
    Say you have a platform, and on that platform you have an object the weighs 10,000lbs (ignore the weight of the platform itself). If the platform is only allowed to move vertically 6” before it must stop moving, how much upward force would be needed to throw the 10,000lb object 12” in the air? Also ignore air resistance.
     
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  3. Oct 31, 2018 #2

    berkeman

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    Is this problem for schoolwork? If not, what is the application? How do you "throw" a 10,000 pound object?
     
  4. Oct 31, 2018 #3
    It was a discussion that I saw recently but I changed the numbers. Obviously you can’t personally throw it but say you had some sort of mechanical device that was doing it, how much force would be needed? I changed the numbers because I wanted to see the effect of higher weight and limited travel distance of the force.
     
  5. Oct 31, 2018 #4

    ZapperZ

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    The question itself is incomplete. For example, what causes the platform to move? Is it supported by a spring? If so, what is the stiffness, or spring constant of the spring? If it is on concrete, why should it even move?

    Secondly, how quickly do I have to lift the object? I can have the Hulk slowly lift it up 12" in the air, or I can put dynamites underneath the object and explode them, causing the object to jump up in the air very quickly. Each one of these scenario will produced different "impulse" and will cause different amount of reaction force being applied to the platform.

    Zz.
     
  6. Oct 31, 2018 #5
    I’ll put a method to it. Say the platform in sitting on a solid object at rest but underneath the platform is hydraulic rams. The stroke of the ram is only 6” so all of the upward force will have to be exerted on the object in the time it takes the ram to move 6” and within that 6” distance. How much would be needed to throw the object a further 12” into the air above the platforms final point (so 18” total above the original starting point, 12” above the platforms ending point).
     
  7. Oct 31, 2018 #6

    sophiecentaur

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    I think this is doable. Work out the speed needed for the 6” throw. Then you need to accelerate the load over the bottom 6” at acceleration x to get that speed in 6”. The force needed will be (x+g)Times the mass.
    There are other approaches but that way gives understandable (?) steps I think.
    I assume you use the maximum lifting distance.
     
  8. Oct 31, 2018 #7

    ZapperZ

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    Again, how "stiff" is this hydraulics? And how fast do you want me to lift this? If I lift it very slowly, the platform will hardly move!

    Zz.
     
  9. Oct 31, 2018 #8
    Don’t really understand your stiff question. When the platform is at its starting position, the hydraulic is not exerting any force on the platform so “zero stiff” I guess? Secondly, as far as speed, that’s part of my question. Obviously if you accelerate slowly the platform won’t be thrown at all. It will rise the 6” with the platform and stop when the platform does. I’m asking what force will be needed to generate the speed needed for a 10,000lb object to overcome the 9.8m/s/s of gravity enough to reach 12” of height before it starts coming back down.
     
  10. Oct 31, 2018 #9
    Yes this is exactly right and what I’m trying to figure out.
     
  11. Oct 31, 2018 #10

    russ_watters

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    @ZapperZ I think the design should follow the requirements, not the other way around. And if the OP doesn't care about the acceleration profile, we might as well assume a constant force is applied. The problem becomes easy then.
     
  12. Oct 31, 2018 #11

    russ_watters

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    This is a fairly straightforward exercise using Newton's classical mechanics. Are you familiar with how the laws of motion work? d=st, v=at, etc?

    To start with, do you know how to calculate the required velocity? Here's a hint: It's the same on the way up as the way down.
     
  13. Oct 31, 2018 #12
    I’m somewhat familiar with them, although it’s been a bit since I practiced it. I’m assuming it’s higher than 9.8m/s? I can’t really remember.
     
  14. Oct 31, 2018 #13

    russ_watters

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    You'll need to actually look up equations and calculate stuff here. Guessing or hoping someone gives you the answer isn't going to work. If you know the relationships between acceleration, speed and distance, post the equations, start manipulating them and see if you can work it out. We'll nudge you if necessary.
     
  15. Oct 31, 2018 #14

    PeroK

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    The first task is to calculate how fast the object must be travelling to go 12" (which is about 30cm in new money) into the air.

    Does the equation ##u^2 = 2gh## look familiar?
     
  16. Oct 31, 2018 #15
    No I don’t remember that one. Let me try that. I got 13.372mph..
     
  17. Nov 1, 2018 #16

    PeroK

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    That doesn't look right. You can in fact, solve this without any calculations.

    If you assume a constant acceleration for the first phase, where the object is accelerated through a distance ##d##, then we also have:

    ##u^2 = 2ad##

    Where ##a## is the required acceleration.

    You can then use the relationship between ##d## and ##h## to get the g-force required. No numbers required.
     
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