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Conservation of energy: Crane raising a mass

  1. Mar 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Mechanical crane raises 225kg at a rate of 0.031m/s^2. from a rest to a speed of 0.5m/s over a distance of 4m. Frictional resistance is 112N.
    m1=225kg
    a1=0.031m/s^2
    u1=0m/s
    v1=0.5m/s
    s=4m
    Fr=112N
    A. Work input from the motor
    B. Tension in the lifting cable
    C. Max power developed by the motor.
    D. Using D'Alembert's principle, solve a) b) c)

    2. Relevant equations
    Wd = ΔPE + ΔKE + Wf (Wf = Work done to overcome friction)
    Fnet = ma
    Wd = Fs
    Weight =m(a+g)
    PE = mgh
    KE = mv^2/2
    D'Alembert's principle; ΣF = 0
    Mechanical Power P = Energy used/Time taken P = FV
    v=u+at
    t=(v-u)/a
    Wd = Work done

    3. The attempt at a solution
    PEi = mgh = 225x9.81x4 = 8829J (incorrect)
    PEi = mgh = 225x9.81x0 = 0J
    PEf = 0 (as the energy is transformed to KE) (incorrect)
    PEf = mgh = 225x9.81x4 = 8829J
    KEi = 0 (starts from a rest)
    KEf = mv^2/2 = 225(0.5x0.5)/2 = 28.13J
    Wd to overcome friction during motion: Wd = Fs = 112x4 = 448J
    Wd (motor) = (PEf - PEi) + (KEf - KEi) + Wf = (8829 - 0)+(28.13-0)+448 = 8352.87
    Therefore the engine Wd would be 8352.87J

    B) Tension in the rope: Tension = weight + Fnet = (ma)+(mg)
    In this case the acceleration of the mass is opposite the acceleration of gravity therefore the formula would be. This is assuming Tension during the acceleration of the mass upwards.
    Tension =m(a+g)
    =(225x0.031)+(225x9.81)
    =2214.23N force of tension on the rope

    c.)
    P=Fv
    =(rope tension + Ff)0.031 = (2214.23+112)0.031 = 72.11W

    D'Alemberts principle states that the total forces must add upto 0. Im not sure how I could show this.

    To me the power output seems to low and I feel like im missing something on energy when calculating the work done, any help would be mucho appreciated sorry about the math grammar.
     
    Last edited: Mar 28, 2017
  2. jcsd
  3. Mar 28, 2017 #2
    Didn't the problem state that the crane "raises" the mass? Then why would the potential energy be less when it is higher?
     
  4. Mar 28, 2017 #3
    It does. I see your point. So I would calculate height as zero for initial PE and PEf would be calculated at the final height destination of the mass in this situation.
     
  5. Mar 28, 2017 #4
    Thanks for the prompt reply Tom
     
  6. Mar 28, 2017 #5
    Yes, I would start out with initial potential energy as 0. That would be the simplest, I think.
     
  7. Apr 18, 2017 #6
    Can anybody see where I am going wrong in my calculations to find tension in the lifting rope. The question is very vague as it doesn't mention if the tension is to be found while at rest or during acceleration.

    Thanks in advance for help.
     
  8. Apr 18, 2017 #7
    Isn't there a friction force also?
     
  9. Apr 18, 2017 #8
    Yeah the friction force is 112N.
    Ahh yes so that would be added to the make the total Fnet.
    Tension = weight + Fnet = (ma)+(mg)+ Fr
    Tension =(225x0.031)+(225x9.81)+112 = 2326.23N
     
  10. Apr 19, 2017 #9
    Would this look about right for the formula's
    Applied force = (Weight+Ff)+(ma)
    Tension = (Weight+Ff)+applied force

    Sorry about the bad drawing and hand writing lol
    20170419_150141 (Medium).jpg
     
  11. Apr 19, 2017 #10
    As I rethink this problem, I think I am confused by 2 things.
    1) Where does the friction force act? Is the friction force acting on the mass? Let's assume for now that it does.
    2) I'm kind of confused by your force definitions. Your diagram shows three forces acting on the mass: friction force, weight force, and force applied. So I assume that the force applied is acting on the mass at the point where the cable attaches to the mass. Is that true? If so, that makes sense to me. So, then where does tension come in? Is that a force acting on the mass, or is that a force somewhere else?

    Maybe the friction force was not intended to be acting on the mass. Do you know for sure?
     
  12. Apr 19, 2017 #11
    Sorry I have not given all the information required Tom I will give the answers now.
    1.) Frictional force is resistance to the motion of the mass.
    2.)Yes, The applied force is the force acting at the point between the cable and the mass which in effect will raise the mass at the specified acceleration.

    Tension is the force in the lifting cable I imagined Tension as all the forces acting on the rope both vertically and horizontally?
     
  13. Apr 19, 2017 #12
    Sorry for throwing the question out there without the required information.
     
  14. Apr 19, 2017 #13
    So if the friction is acting on the mass, then there is no other friction that the cable experiences, or in a pulley that the cable moves around. So then the tension in the cable (at any point in the cable) is the same as the applied force. So then the motor has to do work according to that tension. So I think it is really unnecessary that you have a force called "tension" and another force called "force applied". In this problem, I think my preference would be to use "Tension" rather than "force applied". That's just my preference though.

    Now let's say you had a pulley that had friction. In that case, the tension in the rope would be different on one side of the pulley as it would on the other side.
     
  15. Apr 19, 2017 #14
    Ok so take out the force applied and stick with the tension force. And that would be the result of the tension in the rope between the mass and the pulley?

    A question I have is the equation F=ma is used to find the force required to pulley the mass upwards? I was just wondering as it is such a small force compared to the forces pushing against the mass.
     
  16. Apr 19, 2017 #15
    F = ma should actually be ΣF = ma. The sum of the forces is proportional to the acceleration. Let's forget the friction force for a moment. Let's say you have an enormously large weight (w) connected to a cable. Now if the tension on that cable is also enormously large, but just slightly larger than the weight, then the sum of the forces will be very small. So you will have a very small net force acting on a very large mass, which results in a very small acceleration. Does that make sense?

    Here are some numbers as an example. Mass = 1000 kg. So if g = 10 m/s^2, then the weight is 10,000 N. Now if the tension in the cable happens to be 10,001 N, then, because the tension force is greater than the weight force, the object will have an acceleration upward. However, because the difference is very small, then the acceleration will also be very small. ΣF = 10001 - 10000 = 1 N.
    And ΣF = ma so a = ΣF/m = 1/1000 = 0.001 m/s^2.
     
  17. Apr 20, 2017 #16
    Thats makes sense so ΣF is the net force which is the force proportional to the acceleration of the mass.
    Tension in the rope is ΣF + weight this will show whether the mass is being lifted or lowered and by how much force.
    ΣF = ma = 225kg x 0.031m/s^2 = 6.89N
    Tension = ΣF + w = 6.89N + 2207.25N = 2214.14N

    Thanks Tom very helpful the way you describe these situations makes it very easy to understand for myself.

    So the resistance force to movement of the mass this would be also acting on the rope through the mass and could be added into the tension equation as:
    Tension = ΣF + w + Ff = 6.89N + 2207.25N + 112N = 2326.14N
     
  18. Apr 20, 2017 #17
    Correct.
     
  19. Apr 23, 2017 #18
    A.) I'm not sure how the equation ΣF-ma=0 could apply to a problem using energy. my closest guess which is basically using the conservation of energy which is the energy input minus the energy output equals 0. This is really confusing for me as I always thought D'Alembert's principle referred to forces only.

    B.) I can easily prove ΣF-ma=0
    using the previously calculated numbers.
    6.89N - (225kg x 0.031m/s^2) = 0

    C.) Again I am confused how proving the forces add upto 0 can help to find the power output. I feel like I understand D'Alemberts principle and that it is used to convert a dynamic problem to a static by proving their is an equal and opposite force in equilibrium to all forces.

    Thanks again for any help and big up Tom who has been giving a helping hand throughout my problem solving.
     
    Last edited: Apr 23, 2017
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