Question 1 of 6: Question 1 (paper Question I): Equivalent force-couple system at the vise
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3 (Statics and Dynamics) — 2016-May. 3 hours, closed book (one 8.5″×11″ formula sheet permitted). Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. Every printed question is solved below (all 6), even though the exam instructions require only 4.
Given. The part is clamped at the origin O; the drilling point P is located, from the three labelled dimensions on Figure 1, at $r_{P/O}=\{170,\,-100,\,-200\}\text{ mm}$ from O. The drill exerts a force of magnitude 40 N along its own axis and a couple of magnitude 3 N·m about that same axis, with the angles 36° and 46° on the figure locating the axis: 46° is the angle the drill axis makes below the horizontal (x–z) plane, and 36° is the angle between the horizontal projection of the axis and the x-axis.
Given data
Quantity
Value
Position of P relative to O, $r_{P/O}$
170 i − 100 j − 200 k (mm)
Force magnitude, F
40 N
Couple magnitude, M
3 N·m
Dip below horizontal plane
46°
Azimuth of horizontal projection from x-axis
36°
Find. The equivalent force-couple system $\{\mathbf{F}_R,\ \mathbf{M}_{R,O}\}$ exerted by the vise at O.
Fig. 1 – axes at the vise origin O, position vector r to the drilling point P, and the resulting equivalent force-couple system reported at O.
Check: the position vector and the two locating angles are taken directly from the three labelled dimensions (170 mm, 200 mm, 100 mm → the three axis components of r) and the two labelled angles (36°, 46° → azimuth and dip of the drill axis), using the standard "azimuth-then-dip" convention for such figures. The couple is taken collinear with F (a drill's cutting torque acts about its own thrust axis); "clockwise viewed from the lower right" (looking from the drill shank toward the part, i.e. along +F) places M in the same sense as F by the right-hand rule. This reading is internally consistent and is the standard textbook convention for this class of figure, but the printed sketch leaves some doubt — a grader working from the original exam booklet should confirm the sign of the 200 mm (z) component against the booklet's own axis figure.
Approach. Resolve F into Cartesian components from its two locating angles, form the couple vector M parallel to F with the stated sense, then combine them with the moment of F about O using the principle of equivalent force-couple systems: $\mathbf{F}_R=\mathbf{F}$ and $\mathbf{M}_{R,O}=\mathbf{r}_{P/O}\times\mathbf{F}+\mathbf{M}$.
Resolve F from its two angles. With dip $\theta=46^{\circ}$ below the horizontal plane and azimuth $\phi=36^{\circ}$ of the horizontal projection from the x-axis (rotating toward $-z$):
$$F_x=F\sin\theta\cos\phi,\qquad F_y=-F\cos\theta,\qquad F_z=-F\sin\theta\sin\phi$$
Substituting $F=40\text{ N}$, $\theta=46^{\circ}$, $\phi=36^{\circ}$:
$$\mathbf{F}=\{23.28\,\mathbf{i}-27.79\,\mathbf{j}-16.91\,\mathbf{k}\}\text{ N}$$
Form the couple vector. The couple acts about the drill's own axis with the same sense as F (Step 1's unit vector $\hat{\mathbf{u}}=\mathbf{F}/40$):
$$\mathbf{M}=3\,\hat{\mathbf{u}}=\{1.745\,\mathbf{i}-2.084\,\mathbf{j}-1.268\,\mathbf{k}\}\text{ N}\cdot\text{m}$$
Moment of F about O. With $\mathbf{r}_{P/O}=\{0.170,-0.100,-0.200\}\text{ m}$,
$$\mathbf{r}\times\mathbf{F}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0.170&-0.100&-0.200\\23.28&-27.79&-16.91\end{vmatrix}=\{-3.865\,\mathbf{i}-1.780\,\mathbf{j}-2.397\,\mathbf{k}\}\text{ N}\cdot\text{m}$$
Superpose the couple. $\mathbf{M}_{R,O}=\mathbf{r}\times\mathbf{F}+\mathbf{M}$:
$$\boxed{\mathbf{M}_{R,O}=\{-2.12\,\mathbf{i}-3.86\,\mathbf{j}-3.66\,\mathbf{k}\}\text{ N}\cdot\text{m}},\qquad |\mathbf{M}_{R,O}|=5.73\text{ N}\cdot\text{m}$$
and the resultant force is simply $\boxed{\mathbf{F}_R=\{23.28\,\mathbf{i}-27.79\,\mathbf{j}-16.91\,\mathbf{k}\}\text{ N}}$ (a couple has no effect on the resultant force).
