91Ó°ÊÓ

The first condition of equilibrium is that there must be no net external forces acting on the body, and the second condition is that there must be no net external torques from external forces.

Here we have given that A gate is$4.00\text{″¾}$wide and$2.00\text{″¾}$high.

The weight is$700\text{ N}$.

Also, it makes an angle of $30.0°$ at point C.

(a)

The summation of torque at point $B$ is zero due to the equilibrium. So we have,

$\begin{array}{c}∑{\mathrm{Ï„}}_{\mathrm{B}}=0\\ \left(4\text{″¾}\right)(\mathrm{Tsin}30°)+\left(2\text{″¾}\right)(\mathrm{Tcos}30°)−\left(2\text{″¾}\right)\mathrm{W}=0\\ \mathrm{T}=0.535\text{ }\mathrm{W}\\ \mathrm{T}=0.535\left(700\text{ N}\right)\\ \mathrm{T}=374.5\text{ N}\end{array}$

So, the tension in the wire CD $374.5\text{ N}$.

(b)

The gate moves horizontally with acceleration${\mathrm{a}}_{\mathrm{x}}=0$and the summation of horizontal force by the first condition of equilibrium is zero. So we get,

$\begin{array}{c}∑{\mathrm{F}}_{\mathrm{B}}=0\\ {\mathrm{F}}_{\mathrm{h}}−(\mathrm{Tcos}30°)=0\\ {\mathrm{F}}_{\mathrm{h}}=\mathrm{Tcos}30°\\ {\mathrm{F}}_{\mathrm{h}}=\left(374.5\text{ N}\right)\cos 30°\\ {\mathrm{F}}_{\mathrm{h}}=324.3\text{ N}\end{array}$

The magnitude of the horizontal component of the force at hinge B is $324.3\text{ N}$.

(c)

Here there are two vertical forces exerted on the gate, the vertical force at a point $\mathrm{A}$ is ${\mathrm{F}}_{\mathrm{Av}}$ and the vertical force at a point $B$ is $F_{Bv}$ and the summation of the vertical force is zero (by the first condition of equilibrium)

$\begin{array}{c}∑{\mathrm{F}}_{\mathrm{y}}=0\\ {\mathrm{F}}_{\mathrm{hinge}}+\mathrm{Tsin}30°−\mathrm{W}=0\\ {\mathrm{F}}_{\mathrm{hinge}}=\mathrm{W}−\mathrm{Tsin}30°\\ {\mathrm{F}}_{\mathrm{hinge}}=\left(700\text{ N}\right)−\left(374.5\text{ N}\right)\sin 30°\\ {\mathrm{F}}_{\mathrm{hinge}}=512.7\text{ N}\end{array}$

the combined vertical force exerted by hinges A and B is $512.7\text{ N}$