Question 4.SP.9: A uniform pipe cover of radius r = 240 mm and mass 30 kg is ...
A uniform pipe cover of radius r = 240 mm and mass 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
\pmb{T}=T\frac{\overrightarrow{DC}}{DC}=-\frac{6}{7}T\pmb{i}+\frac{3}{7}T\pmb{j}-\frac{2}{7}T\pmb{k}ANALYSIS:
Equilibrium Equations. The forces acting on the pipe cover form a system equivalent to zero. Thus,
\begin{aligned}&\Sigma \pmb{F}=0: A_x \pmb{i}+A_y \pmb{j}+A_z \pmb{k}+B_x \pmb{i}+B_y \pmb{j}+\pmb{T}-(294 N) \pmb{j}=0 (1) \\&\left(A_x+B_x-\frac{6}{7} T\right) \pmb{i}+\left(A_y+B_y+\frac{3}{7} T-294 N\right) \pmb{j}+\left(A_z-\frac{2}{7} T\right) \pmb{k}=0\end{aligned} \begin{aligned}&\Sigma \pmb{M}_B=\Sigma(\pmb{r} \times \pmb{F})=0: \\&2 r \pmb{k} \times\left(A_x \pmb{i}+A_y \pmb{j}+A_z \pmb{k}\right)+(2 r \pmb{i}+r \pmb{k}) \times\left(-\frac{6}{7} T \pmb{i}+\frac{3}{7} T \pmb{j}-\frac{2}{7} T \pmb{k}\right)+(r \pmb{i}+r \pmb{k}) \times(-294 N) \pmb{j}=0\\&\left(-2 A_y-\frac{3}{7} T+294 N\right) r \pmb{i}+\left(2 A_x-\frac{2}{7} T\right) r \pmb{j}+\left(\frac{6}{7} T-294 N\right) r \pmb{k}=0 (2)\end{aligned}Setting the coefficients of the unit vectors equal to zero in Eq. (2) gives three scalar equations, which yield
A_x=+49.0 N \quad A_y=+73.5 NT = 343 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 N \quad B_x=+245 N \quad B_y=+73.5 NThe 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) \quad \Sigma M_z=0: \frac{3}{7} T(0.48 m)-(294 N)(0.24 m)=0T = 343 N