91影视

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 iswide andhigh.

The weight is 700 N .

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

(a)

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

$\begin{array}{r}鈭�{蟿}_{\mathrm{B}}=0\\ \left(4\mathrm{m}\right)\left(\mathrm{Tsin}鈦�{30}^{鈭�}\right)+\left(2\mathrm{m}\right)\left(\mathrm{Tcos}鈦�{30}^{鈭�}\right)鈭�\left(2\mathrm{m}\right)\mathrm{W}=0\\ \mathrm{T}=0.535\mathrm{W}\\ \mathrm{T}=0.535\left(700\mathrm{N}\right)\\ \mathrm{T}=374.5\mathrm{N}\end{array}$

So, the tension in the wire CD 374.5 N .

(b)

The gate moves horizontally with acceleration $a_x=0$and the summation of horizontal force by the first condition of equilibrium is zero. So we get,

$\begin{array}{r}鈭�F_B=0\\ F_h鈭�\left(T\cos 鈦�{30}^{鈭�}\right)=0\\ F_h=T\cos 鈦�{30}^{鈭�}\\ F_h=\left(374.5\mathrm{N}\right)\cos 鈦�{30}^{鈭�}\\ F_h=324.3\mathrm{N}\end{array}$

The magnitude of the horizontal component of the force at hinge B is 324.3 N .

(c)

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

role="math" localid="1668139394128" $\begin{array}{r}\underset{\mathrm{y}}{鈭�}鈥�=0\\ {\mathrm{F}}_{\text{hinge}}+\mathrm{Tsin}鈦�{30}^{鈭�}鈭�\mathrm{W}=0\\ {\mathrm{F}}_{\text{hinge}}=\mathrm{W}鈭�\mathrm{Tsin}鈦�{30}^{鈭�}\\ {\mathrm{F}}_{\text{hinge}}=\left(700\mathrm{N}\right)鈭�\left(374.5\mathrm{N}\right)\sin 鈦�{30}^{鈭�}\\ {\mathrm{F}}_{\text{hinge}}=512.7\mathrm{N}\end{array}$

the combined vertical force exerted by hinges A and B is 512.7 N