Question 4.SP.9: A uniform pipe cover of radius r = 240 mm and mass m = 30 kg......
A uniform pipe cover of radius r = 240 mm and mass m = 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B.
STRATEGY: Draw a free-body diagram with the coordinate axes shown (Fig. 1) and express the unknown cable tension as a Cartesian vector. Then, apply the equilibrium equations to determine this tension and the support reactions.
MODELING:
Free-Body Diagram. The forces acting on the free body include its weight, which is
W = −mgj = −(30 kg)(9.81 m/s²)j = −(294 N)j
The reactions involve six unknowns: the magnitude of the force T exerted by the cable, three force components at hinge A, and two at hinge B. Express the components of T in terms of the unknown magnitude T by resolving the vector \overrightarrow{DC} into rectangular components:
\overrightarrow{DC} = − (480 mm)i + (240 mm)j − (160 mm)k DC = 560 mm
\mathbf{T}=T{\frac{\overrightarrow{D C}}{D C}}=-{\frac{6}{7}}\,\mathrm{T}{\mathbf{i}}+{\frac{3}{7}}\,T\mathbf{j}-{\frac{2}{7}}\,\mathrm{T}{\mathbf{k}}
ANALYSIS:
Equilibrium Equations. The forces acting on the pipe cover form a system equivalent to zero. Thus,
ΣF = 0: A_{x}{{\mathbf{i}}}+A_{y}{\mathbf{j}}+A_{z}{\mathbf{k}}+B_{x}{\mathbf{i}}+B_{y}{\mathbf{j}}+\mathbf{T}-(294\,{\mathrm{N}}){\mathbf{j}}=0
(A_{x}+B_{x}-{\frac{6}{7}}T){{\mathbf{i}}}+(A_{y}+B_{y}+{\frac{3}{7}}T-294\,{\mathrm{N}}){\mathbf{j}}+(A_{z}-{\frac{2}{7}}T){\mathbf{k}}=0 (1)
ΣM_{B} = Σ(r × F) = 0:
2r\mathbf{k}\times(A_{x}\mathbf{i}+A_{y}\mathbf{j}+A_{z}\mathbf{k})+\left(2r \mathbf{i}+r\mathbf{k}\right)\times\left(-{\frac{6}{7}}T\mathbf{i}+{\frac{3}{7}}T\mathbf{j}-{\frac{2}{7}}T\mathbf{k}\right)+ (r \mathbf{i} + r \mathbf{k}) × (−294 N)\mathbf{j} = 0(-2A_{y}-{\frac{3}{7}}T+294\,\mathrm{N})r\mathbf{i}+(2A_{x}-{\frac{2}{7}}T)r\mathbf{j}+({\frac{6}{7}}T-294\,\mathrm{N})r\mathbf{k}=0 (2)
Setting the coefficients of the unit vectors equal to zero in Eq. (2) gives three scalar equations, which yield
A_{x}=+49.0\,\mathrm{N}\qquad A_{y}=+73.5\,\mathrm{N}\qquad T=343\,\mathrm{N} ◂
Setting the coefficients of the unit vectors equal to zero in Eq. (1) produces three more scalar equations. After substituting the values of T, A_{x},~and~A_{y} into these equations, you obtain
A_{z}=+98.0\,\mathrm{N}\qquad\quad B_{x}=+245\,\mathrm{N}\qquad\quad B_{y}=+73.5\,\mathrm{N}
The reactions at A and B are therefore
A = +(49.0 N)i + (73.5 N)j + (98.0 N)k ◂
B = +(245 N)i + (73.5 N)j ◂
REFLECT and THINK: As a check, you can determine the tension in the cable using a scalar analysis. Assigning signs by the right-hand rule (rhr), we have
(+rhr) \Sigma M_{z}= 0 :\qquad\frac{3}{7}T(0.48~\mathrm{m})-(294~\mathrm{N})(0.24~\mathrm{m})=0\quad\quad T=343~\mathrm{N} ◂
