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Help with 2 problems

  1. Jun 27, 2004 #1
    I seem to be stuck yet again on 2 different problems.


    A claw hammer is used to pull a nail out of a board. The nail is at an angle of 60.0 degrees to the board, and a force F1 of magnitude 500 N applied to the nail is required to pull it from the board. The hammer head contacts the board at point A, which is 0.080 m from where the nail enters the board. A horizontal force F2 is applied to the hammer handle at a distance of 0.300 m above the board.

    What magnitude of force is required to apply the required 500-N force to the nail? (You can ignore the weight of the hammer.)

    I started by summing the torques about point A which gave me:

    [tex] \sum\tau = F_2 * .3 + .08*F_1*tan60 = I\alpha [/tex]

    but, since the weight of the hammer isn't given, you're not able to find I, even if I could find alpha, I would still be an unknown. Tried summing force in the X and Y but that adds more variables.

    The next one is:

    A mass of 11.7 kg, fastened to the end of an aluminum wire with an unstretched length of 0.540m , is whirled in a vertical circle with a constant angular speed of 117 rev/min. The cross-sectional area of the wire is 1.50×10-2 cm^2

    Calculate the elongation of the wire when the mass is at the lowest point of the path. Take the Young's Modulus of aluminum to be 7×10^10 Pa and the free fall acceleration to be 9.80 .m/s^2.

    First I calculated omega, (117 Rev/ 60secs) *(2pi/1rev) = 12.25 rad/s

    I need to find the radius next in order to find out the downward velocity to find the total downward force, at least I think.

    I thought maybe the cross-sectiona area = pi * r^2, but not sure if thats right. Once I find the downward velocity, I was planning of finding the stress and strain to find the rest.

    Anyone able to tell me if my methods are sound? and if so, what am I missing?
  2. jcsd
  3. Jun 27, 2004 #2


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    Staff Emeritus
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    The problem says "you can ignore the mass of the hammer". You don't need to find I and you don't need to find α. Those are not asked. What you can do is use conservation of energy: If you were to apply 500 N as required, and the nail moved, say, dx meters, you would have applied 500dx Joules of energy. Now, calculate how much the handle of the hammer moves (call it dy) if the nail moves dx meters (that's a pretty straight forward 'similar triangles- proportion' problem). Since Fdy= 500dx, once you know dy (it will be a number times dx), you can calculate F.

    If the wire is rotated in a vertical direction, the force on it will be its weight (mass times 9.8) plus the centripetal force necessary to move it in a circle. Then use Young's modulus to find the stretch. Yes, the cross-sectional area is given by [pi]r2. You are told that it is 1.50x102 when the wire is not being stretched.
  4. Jun 28, 2004 #3
    Sorry, still have a few questions, even with what you said I'm a bit confused. Here is the picture I'm using, just for reference purposes.


    What you're saying is dx is in the direction of [tex] F_1 [/tex] which is equal to 500N. and dy is in the direction of [tex] F_2 [/tex]. Which makes since since W = F*D, so W = 500dx, that is equal to the work done by [tex] F_2 [/tex]. So you put Fdy = 500dx. From there though everything I've done to try and solve for dy in terms of dx has fallen apart on me. Only way I can think to solve for dy in terms of dx is use the 500Ndx* cos60 = dy, but that doesn't seem to work.

    Sorry, always been bad with trig. stuff.

    Also, for the second question. How am I supposed to know if the wire is whirled with the fall or against it? If its with the fall that would lessen the downward force, where as if it was with it it'd add to the total downward force.
    Last edited: Jun 28, 2004
  5. Jun 28, 2004 #4
    Well got the hammer one out of the way thanks to the help of Luc in the irc chat. However, the elongation one is still an issue. Here is a check to see what I'm doing:

    cross sectional area = [tex] 1.5 * 10^{-2} cm = 1.5 * 10^{-4} m = \pi * r^2 [/tex]

    solve for r to get, r = .0069m

    [tex] Force = mass * radius * \omega^2 [/tex]
    [tex] \omega = (117rev/min* 2\pi)/(60 secs) = 12.25 rad/s [/tex]
    therefore, 11.7kg * .0069m * (12.25 rad/s)^2 = 12.11N

    that part doesn't make sense, cause the force becomes so small. Any advice at this point?
    Last edited: Jun 28, 2004
  6. Jun 28, 2004 #5

    Doc Al

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    Staff: Mentor

    the hammer is a lever

    Treat the hammer as a lever. The torque applied by F2 equals the torque applied by the nail. The pivot point is where the hammer meets the wood, point A. So: F2*(perpendicular distance to pivot point) = F1*(perpendicular distance to pivot point).
  7. Jun 28, 2004 #6

    Doc Al

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    Find the tension

    The first step is to find the tension in the wire. Apply Newton's second law to the mass. Note that there are two forces acting on the mass.

    Once you find the tension, use it to find the stress (Force/Area) on the wire. Then use Young's modulus to find ΔL.
  8. Jun 28, 2004 #7
    Okay here is what I got. The big problem I had was I couldn't visualize this problem, but I can now and it's alot easier.

    When the object is at the top of the circle we get the following:

    [tex] \omega = 117 rev/min * 2\pi rad/rev * 1/60 min/secs = 12.25 rad/s[/tex]

    [tex] \sum F_y = T - mass * radius * \omega^2 - mass * g = 0 [/tex]


    [tex] T = 1062.75 [/tex]

    If I do the same for the force at the bottom I get T = 833.43

    I then use those as my force in the following equation:

    Y = Force * length/(extension * cross-sectional area)

    on top:
    e = 1062.75* .54/(7*10^10*1.5*10^-4)

    on bottom:

    e = 833.43*.54/(7*10^10*1.5*10^-4)

    I tried that as the answer and it gets returned as wrong.

    Anyone see where I messed up at?
  9. Jun 29, 2004 #8
    (1.5x10^-4 cm^2) * (10^-2m/1cm)^2 = 1.5x10^-8 m^2

  10. Jun 29, 2004 #9

    Doc Al

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    You are mixing up the top and bottom of the circle. At the top of the circle: [itex]T + mg = m\omega^2 r[/itex]; at the bottom of the circle: [itex]T - mg = m\omega^2 r[/itex].
  11. Jun 29, 2004 #10
    You were right, not sure why I had them switched when I typed it, they were right for me on my paper.

    So [tex] T_{top} = 833.43N [/tex]
    [tex] T_{bottom} = 1062.75N [/tex]

    also using cookies correction for the square I get the new equation of:

    [tex] e_{top} = ((833.43N)* .54m)/(7*(10^{10})Pa*1.5*10^{-8} m^2) = .429m [/tex]


    [tex] e_{bottom} = ((1062.75N)* .54m)/(7*(10^{10})Pa*1.5*10^{-8} m^2) = .546m [/tex]

    both answers give me the message that I'm off by an additive constant, but I'm not sure where.

    This is how I got omega:

    [tex] \omega = (117rev/min* 2\pi)/(60 secs) = 12.25 rad/s [/tex]

    and then I used the equation:

    Y = Force & original lentgh/(extension length * area), where Y is the youngs modulous which is given as 7*10^10 Pa

    Anyone got any ideas? Need to finish this by midnight tonight, 6/29/04

    Last edited: Jun 29, 2004
  12. Jun 29, 2004 #11

    Doc Al

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    wrong area

    Check your conversion of the area units from [itex]{cm}^2[/itex] to [itex]m^2[/itex].
  13. Jun 29, 2004 #12
    I rechecked my math and found a slight error.

    1cm^2 = .0001 m^2

    So that would be 10^-6 instead of -8.

    So my two equations would be:

    [tex] e_{top} = ((833.43N)* .54m)/(7*(10^{10})Pa*1.5*10^{-6} m^2) = 4.29 * 10^{-3} m[/tex]

    [tex] e_{bottom} = ((1062.75N)* .54m)/(7*(10^{10})Pa*1.5*10^{-6} m^2) = 5.46 * 10^{-3} m [/tex]

    oddly enough, the [tex] e_{bottom} [/tex] gives me an error that might trig might be off, while the other one still says i'm off by some constant.

    I just can't see what else might be off.
  14. Jun 30, 2004 #13
    Bah! after working on it more I found out the website I do my homework on, was rejecting my answers in the form of 4.5*10^-3, once I wrote it out by hand it accepted it fine, I had the right answer for quite some time!

    Thanks for all the help. I really had trouble because I had a really different and really hard image in my head the entire time. Once I got the right image things began to fall in place!

    Thanks again,

  15. Oct 14, 2010 #14
    thanks so much. it turns out the problem I was having was with the radius. A stupid misread direction can lead to a lot of trouble...

    Thanks again!
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