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Homework Help: Part of the floor of a workshop.help

  1. Nov 30, 2012 #1
    Part of the floor of a workshop........help!!

    Part of the floor of a workshop is inclined at 10 degrees to the horizontal. This is to allow the safe...
    ...storage of steel cylinders. A cylinder of mass 7000 kg is stored as shown in the diagram (c the link below). The magnitudes of the forces exerted on the cylinder by the wall and the floor are Rw and Rf newtons respectively. Find the values of Rw and Rf.

    If the sloping part of the floor was inclined at alpha degrees (0<alpha<45) instead of at 10 degrees, show that

    a) Rw<70000<Rf, whatever the value of alpha.

    b) Rw and Rf both increase with alpha.

  2. jcsd
  3. Nov 30, 2012 #2
    Re: Part of the floor of a workshop........help!!

    What have you done to solve this problem?
    Have you drawn a free-body diagram for the cylinder?
  4. Nov 30, 2012 #3
    Re: Part of the floor of a workshop........help!!

    i did this:
    R (II to slope) :
    Rwcos10 = 70000sin10 ---- Rw = 12342.9 N
    R (perpendicular to slope) :
    Rf = 70000cos10 + 12342.9sin10 ----- Rf = 71079.9 N

    I am able to get Rw and Rf on the cylinder. But I don't know how to approach mathematically expressing the second part of the question (the proof part). Parts (a) and (b) of the second part. Could you please help?
  5. Nov 30, 2012 #4
    Re: Part of the floor of a workshop........help!!

    I think the best way of doing this is to go back to your free-body diagram and write the x and y components out in symbols - don't put in any numbers at the moment.

    Also, it would be easier to take the x-axis along the dotted line in your diagram.

    When you do this, the expressions for Rw and Rf will each involve 1 trig function. By looking at the values of the trig functions for the given angle range 0<alpha<45, you should be able to get the answer.

    If you have problems, show us your expressions for the components.
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