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Find force required to move pole with static friction

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data

    The uniform 14-ft pole weighs 150 lbs and is supported as shown. Calculate the force P required to move the pole if the coefficient of static friction for each contact location is 0.40.

    (Sorry for the sideways images!)

    33eonbd.jpg

    2. Relevant equations
    1.) Fmax = μs*N
    2.) ∑M = 0
    3.) ∑Fx = 0
    4.) ∑Fy = 0

    3. The attempt at a solution

    Step 1

    I began by setting up a FBD (shown below). At points A and B I have a Normal force N and friction force F. I placed the weight vector (150 lbs) in the middle of the 14 ft beam (so 7 ft). I calculated the angle closest to point B in order to find the x-direction distance to the weight vector (which I found to be 5.6 ft).

    Step 2

    Next I took the moment at point B, which gave me (-8)(NA) + (-5.6)(-150) - (6)(FA) = 0.

    Based on equation #1 I assumed FA = (.40)(NA), so I used that to find NA = 80.769 lbs

    Step 3
    Using ∑Fy = 0, I found NA + NB - 150 = 0 to find NB = 69.230 lbs

    Step 4
    Fx = 0: ⇒ FA + FB - P = 0, which I found P = 60.

    My book says the answer is 118.5 lbs, so I know I'm way off, but I'm not sure where exactly. I'm thinking I'm off somewhere in my FBD, but could be elsewhere too.

    2qn5pty.jpg
     
  2. jcsd
  3. Apr 28, 2015 #2

    haruspex

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    What is meant by a "normal" force? Which way is the normal force when a point contacts flat or curved surface?
     
  4. Apr 28, 2015 #3
    After thinking about it, I realize that it's force perpendicular to the surface, so I FA direction would need to be perpendicular to the corner it's sitting on? And would the friction force then be perpendicular to the normal force? If that's the case, how would I go about solving for the magnitude of FA and FB without knowing the angle?
     
  5. Apr 28, 2015 #4

    haruspex

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    Yes, except that it doesn't mean anything to say 'perpendicular to the corner'. You mean perpendicular to th surface, so friction is parallel to the pole.
    You know the angle from the given lengths, but there's no point in calculating the angle as such. You will only need trig functions of the angle, and these can be derived directly from the lengths.
    A tip: always work algebraically as far as possible, only plugging in numbers at the final step. It minimises rounding errors and maximises comprehensibility. It often reduces effort too, since some variables might cancel out.
     
  6. Apr 28, 2015 #5
    That got it, thanks again! I see what you mean about working algebraically seeing as I ended up with 118.66 due to rounding errors, but that's definitely the right method.
     
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