Chapter 2.3, Problem 2.62P

Free-Body Diagram

$\begin{array}{l}{T}_{AC}=T_{BC}=800\text{N}\\ AC=BC=\sqrt{\left(h^2+d^2\right)}\\ \sum F_y=0:2\left(800\text{N}\right)\frac{h}{\sqrt{\left(h^2+d^2\right)}}-P=0\\ 800=\frac{P}{2}\sqrt{1+{\left(\frac{d}{h}\right)}^2}\end{array}$

Data: P = 200 N, d = 600 mm and solving for h

$800\text{N}=\frac{200\text{N}}{2}\sqrt{1+{\left(\frac{600\text{mm}}{h}\right)}^2}$

h = 75.6 mm

2.62 For W = 800 N, P = 200 N, and d = 600 mm, determine the value of h consistent with equilibrium.

Fig. P2.62