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Homework Help: Hammer/Nail problem involving torque

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Figure P8.58 shows a claw hammer as it is being used to pull a nail out of a horizontal board. If a force of magnitude 150N is exerted horizontally as shown, find a) the force exerted by the hammer claws on the nail and b) the force exerted by the surface at the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail.

    Figure P8.58
    http://img137.imageshack.us/img137/2198/hammerbp2.png [Broken]

    2. Relevant equations
    T = rFSin[tex]\vartheta[/tex]

    3. The attempt at a solution
    Using trig I find that the force the prongs exert on the nail is 300N
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 16, 2008 #2
    you should check your answer.

    the torque caused by the horizontal force is 45. This means that the vertical force exerted is 900.
    Last edited: Mar 16, 2008
  4. Mar 16, 2008 #3
    How did you get your answers?
  5. Mar 16, 2008 #4
  6. Mar 16, 2008 #5
    Why do you do it that way instead of the way I did it? Can you explain that please?
  7. Mar 16, 2008 #6
    Well, how did you do it? The vertical force exerted on the nail is 900N. The nail needs to exert a component that would equal 900N. The toher component turns up as the contact force 9well part of, not forgeting the weight of the hammer).
    Last edited: Mar 16, 2008
  8. Mar 16, 2008 #7
    This is how I did it:

    My diagram looks like this
    http://img219.imageshack.us/img219/7730/nailod0.png [Broken]

    Since the 150N is the horizontal component of the force exerted on the nail I used trig to find the force.

    sin30 = 150/F
    F = 150/sin30 = 300
    Last edited by a moderator: May 3, 2017
  9. Mar 16, 2008 #8
    you should do the question via moments.

    Imagine you had a longer handle, wouldn' it be easier to retrieve the nail?
  10. Mar 16, 2008 #9
    Yeah, you're right it would. But I still don't see how you got your numbers :(
  11. Mar 16, 2008 #10
    the torque caused by the hammer is 150 x 0.3.

    if the nail is not moving, then it is countering this torque with its own. So to counter this, it would need to exert a force which has a component that would counter this torque.

    EDIT: i edited my last few posts for clarity
    Last edited: Mar 16, 2008
  12. Mar 16, 2008 #11
    by the way my previous number of 788 was wrong it should be higher than 900
  13. Mar 17, 2008 #12
    I tried it by moments as oerg suggested:

    Taking moments about the point of contact of the hammer and the floor (this eliminates the reaction of the floor).
    In equilibrium:

    moment of effort on hammer =150*0.3 = moment of hammer on nail force= F*(0.05/cos(30 deg))

    solving for F gives 780 N.
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