Question 4.SP.8: A 5 × 8-ft sign of uniform density weighs 270 lb and is supp......

STRATEGY: Draw a free-body diagram of the sign, and express the unknown cable tensions as Cartesian vectors. Then, determine the cable tensions and the reaction at A by writing and solving the equilibrium equations.

MODELING:

Free-Body Diagram. The forces acting on the sign are its weight W = –(270 lb)j and the reactions at A, B, and E (Fig. 1). The reaction at A is a force of unknown direction represented by three unknown components. Because the directions of the forces exerted by the cables are known, these forces involve only one unknown each: specifically, the magnitudes T_{BD}~and~T_{EC}. The total of five unknowns means that the sign is partially constrained. It can rotate freely about the x axis; it is, however, in equilibrium under the given loading, because the equation ΣM_{x} = 0 is satisfied.

ANALYSIS: You can express the components of the forces T_{BD}~and~T_{EC} in terms of the unknown magnitudes T_{BD}~and~T_{EC} as follows:

\overrightarrow{{{B D}}}=-(8\ \mathrm{ft})\mathbf{i}+(4\ \mathrm{ft})\mathbf{j}-(8\ \mathrm{ft})\mathbf{k}\qquad B D=12\ \mathrm{ft}

\overrightarrow{{{E C}}}=-(6\ \mathrm{ft})\mathbf{i}+(3\ \mathrm{ft})\mathbf{j}+(2\ \mathrm{ft})\mathbf{k}\qquad{E C}=7\ \mathrm{ft}

\mathbf{T}_{B D}=T_{B D}{\left({\frac{\overrightarrow{B D}}{B D}}\right)}=T_{B D}(-{\frac{2}{3}}\mathbf{i}+{\frac{1}{3}}\mathbf{j}-{\frac{2}{3}}\mathbf{k})

\mathbf{T}_{EC}=T_{EC}{\left({\frac{\overrightarrow{EC}}{EC}}\right)}=T_{EC}(-{\frac{6}{7}}\mathbf{i}+{\frac{3}{7}}\mathbf{j}+{\frac{2}{7}}\mathbf{k})

Equilibrium Equations. The forces acting on the sign form a system equivalent to zero:

ΣF = 0:                      A_{x}{\mathbf{i}}+A_{y}{{\mathbf{j}}}+A_{z}{\mathbf{k}}+\mathbf{T}_{B D}+\mathbf{T}_{E C}-(270\;\mathrm{{lb}}){{\mathbf{j}}}=0

(A_{x}-{\frac{2}{3}}T_{B D}-{\frac{6}{7}}T_{E C})\mathbf{i}+(A_{y}+{{\frac{1}{3}}}T_{B D}+{\frac{3}{7}}T_{E C}-270\;\mathrm{lb})\mathbf{j}+\,(A_{z}-{\frac{2}{3}}T_{B D}+{\frac{2}{7}}T_{E C})\mathbf{k}=0          (1)

ΣM_{A} = Σ(r × F) = 0:                     (8\mathrm{~ft})\mathbf{i}\times T_{B D}(-{\frac{2}{3}}\mathbf{i}+{\frac{1}{3}}\mathbf{j}{-{\frac{2}{3}}}\mathbf{k})+(6\mathrm{~ft})\mathbf{i}\times T_{E C}(-\frac{6}{7}\mathbf{i}+\frac{3}{7}\mathbf{j}+{\frac{2}{7}}\mathbf{k})+ (4 ft)\mathbf{i} × (−270~lb)\mathbf{j} = 0\\

(2.667T_{B D}+2.571T_{E C}-1080~lb){\bf k}+(5.333T_{B D}-1.714T_{E C}){\bf j}=0      (2)

Setting the coefficients of j and k equal to zero in Eq. (2) yields two scalar equations that can be solved for T_{BD}~and~T_{EC}:

T_{BD} = 101.3 lb              T_{EC} = 315 lb ◂

Setting the coefficients of i, j, and k equal to zero in Eq. (1) produces three more equations, which yield the components of A.

A = + ( 338 lb )i + ( 101.2 lb )j − ( 22.5 lb )k ◂

REFLECT and THINK: Cables can only act in tension, and the free-body diagram and Cartesian vector expressions for the cables were consistent with this. The solution yielded positive results for the cable forces, which confirms that they are in tension and validates the analysis.