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    vikram_gupta11's Avatar
    vikram_gupta11 Posts: 44, Reputation: 1
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    #1

    Apr 30, 2017, 06:44 AM
    Velocity transfer
    Two 100 kilogram balls are mounted on a seesaw and this seesaw is in balanced position .Now I want to know that how much force I should apply on the right side arm of this balanced seesaw so that the left side 100 kilogram weight ball could jump in the air up to 5 centimeter.

    This question is regarding a project and it is not a homework related.
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #2

    Apr 30, 2017, 08:41 AM
    I assume what you mean is you want the 100 Kg ball on the left to fly up off the seesaw when the right side hits the ground and stops - correct? You'll need to tell us the travel distance over which the force on the right side operates - in other words how far does the right side travel from the startng equilibrium point before hitting the ground?
    vikram_gupta11's Avatar
    vikram_gupta11 Posts: 44, Reputation: 1
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    #3

    Apr 30, 2017, 11:16 AM
    The travel distance of right side will be 10 centimeter from the ground.
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #4

    Apr 30, 2017, 11:50 AM
    You will need about 1000 N force applied downward on the right hand side of the seesaw.
    vikram_gupta11's Avatar
    vikram_gupta11 Posts: 44, Reputation: 1
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    #5

    Apr 30, 2017, 12:15 PM
    Dear sir,can you please provide me some equations and formulas regarding your reply.
    ebaines's Avatar
    ebaines Posts: 12,130, Reputation: 1307
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    #6

    May 1, 2017, 05:33 AM
    Use energy principles:

    1. The velocity you want at the instant the seesaw stops is v^2 = 2gh, where h = 0.05 meters (the height the 100 Kg mass needs to rise above the stopping point).

    2. The work put into the system is F times d, where F is the applied force and d is the distance the force is applied over, which is 0.1m.

    3. Conservation of energy:
    Change in kinetic energy of the two masses + change in potential energy of the two masses = work applied.

    Can you take it from here?

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