Question 5.3: A uniform semicircular rod of weight W and radius r is attac...
A uniform semicircular rod of weight W and radius r is attached to a pin at A and rests against a frictionless surface at B. Determine the reactions at A and B.
STRATEGY: The key to solving the problem is finding where the weight W of the rod acts. Since the rod is a simple geometrical shape, you can look in Fig. 5.8 for the location of the wire’s centroid.
MODELING: Draw a free-body diagram of the rod (Fig. 1). The forces acting on the rod are its weight W, which is applied at the center of gravity G (whose position is obtained from Fig. 5.8B); a reaction at A, represented by its components \pmb{\text{A}}_x \text{ and } \pmb{\text{A}}_y; and a horizontal reaction at B.
ANALYSIS:
\begin{matrix} +\circlearrowleft \sum{M_A}=0: && B(2r) -W(\frac{2r}{\pi})=0 \\ && B=+\frac{W}{\pi} && \pmb{\text{B}}=\frac{W}{\pi}\longrightarrow \\ \underrightarrow{+}\sum{F_x}=0: && A_x+B=0 \\ && A_x=-B=-\frac{W}{\pi} && \pmb{\text{A}}_x=\frac{W}{\pi}\longleftarrow \\ +\uparrow \sum{F_y}=0: && A_y-W=0 & &\pmb{\text{A}}_y=W\uparrow \end{matrix}
Adding the two components of the reaction at A (Fig. 2), we have
\begin{matrix} A=\left[W^2+\left(\frac{W}{\pi}\right)^2 \right]^{1/2} && A=W\left(1+\frac{1}{\pi^2}\right)^{1/2} \\ \tan \alpha = \frac{W}{W/\pi}=\pi && \alpha=\tan^{-1}\pi \end{matrix}
The answers can also be expressed as
\pmb{\text{A}}=1.049W\measuredangle 72.3^\circ \quad \quad \quad \quad \quad \pmb{\text{B}}=0.318W\longrightarrow
REFLECT and THINK: Once you know the location of the rod’s center of gravity, the problem is a straightforward application of the concepts in Chapter 4.
