# Mechanics: Manual Drill Eccentricity

1. Sep 28, 2011

### JohanM

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
$\vec{M}=\vec{r}\times\vec{F}$

3. The attempt at a solution

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

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

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

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:

$F_{R}=(-10)\hat{i} + (-30+15)\hat{j} + 0\hat{k}$
$\theta=tan^{-1}(\frac{15}{10})=56.3^{\circ}$

$M_{A}=(10)(0)+(-15 lb)(16 in)+(30 lb)(8 in)$
$=0$

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