Question 1 (paper Question I) — 3D Rod Assembly Supporting a Cylinder (Part A · Statics, equal value): 04-BS-3, December 2019, Q1

Question 1 (paper Question I) — 3D Rod Assembly Supporting a Cylinder (Part A · Statics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rigid rod assembly, weightless. Ball-and-socket support at A. Rod bends: A to C is 2 m, C to D is 1.5 m (vertical, a separate two-force member CD pinned at both ends), C to E is 1 m (perpendicular to A-C), E to F is 1.5 m (parallel to A-C). A 250 N cylinder hangs at F. E is a smooth journal bearing whose axis is along the C-E direction (so it develops no reaction along C-E).

Given data — joint coordinates (m), using local axes u (A→C direction), v (C→E direction), w (vertical)
Point	u	v	w	Role
A	0	0	0	ball-and-socket joint
C	2	0	0	rod bend / CD lower pin
D	2	0	1.5	CD upper pin (2-force member)
E	2	1	0	smooth journal bearing
F	3.5	1	0	250 N cylinder support

Find. The reaction components $A_u,A_v,A_w$ at A, the bearing reaction components $E_u,E_w$ at E, and the force $T_{CD}$ in rod CD.

Figure 1 (source dimensions, local u-v-w axes). Ball-and-socket at A; rod bends at C (branch up to D) and again at E (smooth journal bearing); the 250 N cylinder hangs at F.

Approach. Treat the rod A-C-E-F plus the two-force member CD as one rigid body; write $\Sigma\vec F=0$ and $\Sigma\vec M_A=0$ in Cartesian vector form and solve the resulting 6×6 linear system for the six unknown reaction components.

Position vectors and the CD unit vector. With A as origin, $\vec r_{C}=(2,0,0)$, $\vec r_{D}=(2,0,1.5)$, $\vec r_{E}=(2,1,0)$, $\vec r_{F}=(3.5,1,0)$ m. Rod CD is a two-force member, so its force on the frame at C acts along $$\hat u_{CD}=\frac{\vec r_D-\vec r_C}{|\vec r_D-\vec r_C|}=(0,0,1)$$ Let $T$ be the (tension-positive) force in CD.
Set up the reaction vectors. Ball-and-socket at A: $\vec R_A=(A_u,A_v,A_w)$. Smooth journal bearing at E (no reaction along its own axis, the v-direction): $\vec R_E=(E_u,0,E_w)$. Applied load at F: $\vec P=(0,0,-250)$ N.
Force equilibrium. $$\Sigma\vec F=\vec R_A+\vec R_E+T\hat u_{CD}+\vec P=0$$
Moment equilibrium about A. $$\Sigma\vec M_A=\vec r_E\times\vec R_E+\vec r_C\times(T\hat u_{CD})+\vec r_F\times\vec P=0$$ Expanding the cross products and collecting the $u,v,w$ components gives six scalar equations in the six unknowns $A_u,A_v,A_w,E_u,E_w,T$.
Solve the linear system. Expanding $\vec r_E\times\vec R_E=(E_z,\,-2E_z,\,-E_u)$, $\vec r_C\times(T\hat u_{CD})=(0,\,-2T,\,0)$ and $\vec r_F\times\vec P=(-250,\,875,\,0)$ and summing component-by-component: the $w$-component gives $-E_u=0\Rightarrow E_u=0$; the $u$-component gives $E_w-250=0\Rightarrow E_w=250$ N; the $v$-component then gives $$-2E_w-2T+875=0\ \Rightarrow\ -2(250)-2T+875=0\ \Rightarrow\ \boxed{T_{CD}=187.5\text{ N (tension)}}$$ Back-substituting into the three force equations gives $A_u=0$, $A_v=0$, and $$A_w=250-E_w-T=250-250-187.5=\boxed{-187.5\text{ N}}$$

Final Results — Question 1
Quantity	Value
$A_u$	0 N
$A_v$	0 N
$A_w$	-187.5 N
$E_u$	0 N
$E_w$	250 N
$T_{CD}$	187.5 N (tension)

Topic: Equilibrium of a rigid body in three dimensions (Cartesian vector method).

Key relations: $\Sigma\vec F=0$, $\Sigma\vec M_O=0$ (six independent scalar equations); a ball-and-socket joint supplies 3 unknown force components; a smooth journal bearing supplies 2 (no component and no moment along its own axis); a two-force member's reaction is constrained to lie along the line joining its two pins.

Why this works. A single rigid body in 3-D space has six degrees of freedom, so exactly six independent scalar equilibrium equations are needed and available whenever the supports supply exactly six unknowns (3+2+1 here). Writing every position vector from one common point (A) and using the vector cross product keeps track of every moment arm automatically, which is far less error-prone than resolving each force into components by inspection.

Common pitfall: forgetting that a smooth journal bearing develops no reaction (and no moment) along its own axis, which would over-count the unknowns and make the system unsolvable by statics alone; also mixing up which point a two-force member's force vector points toward (it must point from the joint toward the far pin for a positive/tension sign convention).

Source: Hibbeler, Engineering Mechanics: Statics, Ch. 5 (Equilibrium of a Rigid Body — three dimensions).

Question 1 of 6: Question 1 (paper Question I) — 3D Rod Assembly Supporting a Cylinder (Part A · Statics, equal value)

Question 1 (paper Question I) — 3D Rod Assembly Supporting a Cylinder (Part A · Statics, equal value)