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Two Opposing Forces and Velocity

  1. Nov 3, 2014 #1
    In a crane like mechanism, a 0.25 Kg object is lifted with a pulling force of 100 Newtons. Neglecting friction, find the velocity of the object being lifted.

    ma = F
    0.25kg*9.81m/s2 = 2.45N

    F1 = 100N
    F2 = 2.45N

    My problem here is getting the velocity. Please help!
     
  2. jcsd
  3. Nov 3, 2014 #2

    BvU

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    Hello Frank, and welcome to PF :)

    Unfortunately, help is to no avail in this situation. Either the problem itself, or your rendering of it, lacks the information to find the velocity. You sure this is the exercise, the whole exercise, and nothing but the exercise ? :) again...
     
  4. Nov 3, 2014 #3
    Hello BvU

    Sorry if my the problem was lacking data. And yes I made it myself. :P

    It's actually for my scaled crane project, I was thinking of making a document that will compare the actual and theoretical data to obtain my crane's efficiency.

    What more data should I add to be able to solve the velocity? Well my motor's linear velocity is 10.6029cm/s, say that the hanging object is to travel 45cm.

    Extra**

    Correct me if I'm wrong, this is how I solved for the motor's linear velocity:
    Motor Specs (90rpm, 6V, 2.4A), Pulley diameter is 2.25cm

    ω = 90rpm * 2π rad/rev
    ω = 180π rad/min

    Velocity = ω * (radius of pulley)
    V = 180π * 1.125cm
    V (converted minutes to seconds) = 10.6029 cm/s
     
  5. Nov 3, 2014 #4

    BvU

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    So if the pulley pulls in 11 cm of 'cable' per second, doesn't the load move upwards at 11 cm/s too ?

    However, there are a few more things to consider:
    Motor speed: is this 90 rpm under free running conditions, or is it under full load ?
    And the 100 N: where does that come from ? Again: free running or full load ?

    As you calculated, the 100 N is more than enough to lift the 0.25 kg load.

    Have you learned about Newton's law already ? Fnet = ma ? if the lifting speed is constant (for example this 11 cm/s), then the acceleration is zero, so the net force is zero too: the pulling force and the weight should then be equal (both 2.45 N, in opposite directions).

    As you see: ask a physicist a question and he homes back with no answer and a lot more questions :)
     
  6. Nov 3, 2014 #5
    Yes the linear velocity of the motor is 11 cm/s, but isn't it going to slow down since the motor is pulling 250 grams? That's what I've been trying to figure out. :confused:

    Yes 90rpm is under free running condition.

    It's actually 136 N to be exact, I just rounded it off, obtained from using the Power (Horsepower) = (Tq * Rpm)/5252 formula using 6 volts and 2400ma, those are 4 AA batteries in series. :p Then used the Torque = Force*Radius formula. And I got 136 N, free running condition as well. :)

    Yes I've been there, the equilibrium law..
     
  7. Nov 3, 2014 #6

    BvU

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    Ah, now I see. Stick to SI units if you can: 6V * 2.4 A = 14.4 W. And 14.4W/0.106 m/s is about 136 N.

    The batteries won't provide 1.5 V at 2.4 A (and definitely not for a long time...)
    There's loss in the wires, in the motor, in the pulleys, ...
    I think even your original 100 N is optimistic... but fortunately you only need 2.5, so there's a margin.

    Motor is going to slow down under load. How much depends on its characateristics (google "motor characteristics", or "torque rpm"). But much better: try it out !

    Or leave it all and consider these calculations are for the ideal situation
     
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