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Finding the Force on a bolt from a wrench

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    In an Effort to tighten a bolt, a force F is applied to the end of a wrench that has a length L perpendicular to the wrench. If the bolt itself has a radius of b, how much force is applied to the bolt

    2. Relevant equations
    ∑τ=RFsinθ ΣF=ma ∑T=Iα

    3. The attempt at a solution
    I tried summing the torques
    ∑τ=LF+bF=(I_bolt+I_wrench)α

    but the answer choices are
    a)F
    b)Fb/L
    c)FL/b
    d)FL/(b+L)
    e)bL/F

    I'm not even sure how you would solve this to get anything like that or why there is an F in the answer when we're solving for F. My friend says this is a conceptual engineering problems and that c makes sense since the Force is multiplied by the length of the bolt (they are really looking for torque) but divided by the length of the wrench.
     
  2. jcsd
  3. Apr 12, 2016 #2

    BvU

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    That's not the title of your thread. You sure they aren't out to trick you ?

    And: make a drawing
     
    Last edited: Apr 12, 2016
  4. Apr 12, 2016 #3
    here's a pic drawn in paint
     

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  5. Apr 12, 2016 #4

    BvU

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    L is a bit vague, don't you think ?

    And do you think people mean that when they mention the radius of a bolt ? In the civilized world the M10 bolt has a radius of 5 mm, but the head is a bit bigger !

    Note that there is only one F in your drawing !

    You be the judge: answering the question literally is a). But the force the thread of the bolt exerts on the thread on the inside of the hole (or nut) is c) as your friend sort of indicated. This multiplication factor is the reason wrenches exist :smile:

    (Note that the length of a bolt is in general not the same as the radius...)

    You know the torque from L x F. (angle is 90 degrees). At the thread of the bolt the torque is the same !

    None of the answers has the dimension of torque ( Newton x meter or better: Force x length). All of them have the dimension of a force.
     
  6. Apr 12, 2016 #5
    So are we looking for a ratio of the torques to find the Force?
     
  7. Apr 12, 2016 #6
    Oh a radius of a bolt is the round thing before the threat, is the force applied there?
     

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  8. Apr 12, 2016 #7

    BvU

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    No. That's a specific type of bolt and that radius has little to do with the exercise.

    I meant the radius of the shaft: so D/2 in the drawing below

    bolts.gif

    I tried to explain that there is only one torque and that the ratio of the forces at L and at b is the ratio of b and L : ##\tau = L \times F_L ## and ##\tau = b \times F_b##
     
    Last edited: Apr 12, 2016
  9. Apr 12, 2016 #8
    Ohhh that makes so much sense thanks
     
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