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Mechanics: Manual Drill Eccentricity

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    In order to manually drill a hole, as shown below, it is important to keep the drill perpendicular to the surface. The drill is spun by a 30-lbf force applied downward at the handle at B, while pressure on the bit at A is obtained by applying a 10-lbf force along the x-axis at C. The handle currently lies along the x-z plane and the wall does not supply a reaction moment at A.

    a) Find the other axial components of the force applied at C that are required to keep the bit at A oriented solely along the x-axis.
    b) Find the equivalent force and moment couple at A that results from the applied forces at B and C.
    2. Relevant equations

    3. The attempt at a solution

    My approach was to determine the moments of B and C about point A first.

    [itex]\vec{M}^{F_{B}}_{A}=\left\langle 8,0,-6 \right\rangle \times \left\langle 0,-30,0 \right\rangle [/itex]
    [itex]=\left\langle -180,0,-240 \right\rangle [/itex]

    [itex]\vec{M}^{F_{C}}_{A}=\left\langle 16,0,0 \right\rangle \times \left\langle -10,C_{y}, C_{z} \right\rangle [/itex]
    [itex]=\left\langle 0, -16C_{z}, 16C_{y} \right\rangle [/itex]

    I then assumed that only the x-component of the moments about A should be non-zero, and said that Cz=0 and 16Cy=240, yielding Cy=15 lb.

    However, when I calculate the resultant force and the moment couple at A, the moment couple equals zero:

    [itex]F_{R}=(-10)\hat{i} + (-30+15)\hat{j} + 0\hat{k}[/itex]

    [itex]M_{A}=(10)(0)+(-15 lb)(16 in)+(30 lb)(8 in)[/itex]

    Clearly, the moment couple at point A is not zero, so my approach is wrong...
    What should I change?

    Thanks in advance,
  2. jcsd
  3. Sep 29, 2011 #2


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    Gold Member

    Why can't the moment couple at point A be zero? If it were not zero wouldn't the drill not stay on axis? Sorry if this is an obvious question.
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