91影视

Given that a uniform drawbridge must be held at a $37掳$angle above the horizontal to allow ships to pass underneath. The drawbridge weighs 45000 N and is 14.0 m long. A cable is connected 3.5 m from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place.

Weight of drawbridge, w = 45000 N

Moment of arm for w, ${\mathrm{I}}_1=\frac{14\mathrm{m}}{2}=7\mathbf{m}$

Let T be the tension in the cable when drawbridge is at angle of $37掳$with the horizontal.

Moment of arm for tension T is I = 3.5 m

Torque $蟿=FI$

Where Fis force exerted and l is moment arm.

Apply first condition for equilibrium.

Net torque on the axis of the hinge is zero.

This implies $TI\sin 37掳-w{l}_1\cos 37掳=0$

$\begin{array}{r}T\left(3.5\mathbf{m}\right)\left(\sin 鈦�{37}^{鈭�}\right)=\left(45000\mathbf{N}\right)\left(7\mathbf{m}\right)\left(\cos 鈦�{37}^{鈭�}\right)\\ T=\frac{2\left(45000\mathbf{N}\right)}{\tan 鈦�{37}^{鈭�}}\\ T=1.19脳{10}^5\mathbf{N}\end{array}$

Hence tension in the cable is$1.19脳{10}^5\mathbf{N}$

Let $H_h$and$H_v$ are horizontal and vertical components of force exerted by hinge on the bridge H.

Now, applying second condition of equilibrium,

Net force is zero.

Thus $\underset{}{鈭�}{F}_x=0and\underset{}{鈭�}{F}_y=0$

$鈭�F_x=0$implies $H_h=T=1.19脳{10}^5\mathbf{N}$

$鈭�F_y=0$implies localid="1668070667740" $H_v=w=4.5脳{10}^4\mathbf{N}$

Therefore force exerted is $H=\sqrt{H_h^2+H_v^2}=1.27脳{10}^5\mathit{N}$

$\begin{array}{r}\tan 鈦�胃=\frac{H_v}{H_h}\\ 胃={20.7}^{鈭�}\end{array}$

Thus, hinge force has magnitude $1.27脳{10}^5\mathbf{N}$and is directed at $20.7掳$above horizontal.

Treat bridge as a uniform bar rotating around one end, so

Moment of Inertia, $I=\frac{1}{3}m{L}^2$ where m is mass of bridge and L is length of bridge.

Torque is $蟿=l伪$where $伪$is angular acceleration.

This gives $mg\left(\frac{L}{2}\right)\cos 37掳=\frac{1}{3}m{L}^2伪$

Thus

$\begin{array}{r}伪=\frac{3g\cos 鈦�{37}^{鈭�}}{2L}\\ =\frac{3\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cos 鈦�{37}^{鈭�}}{2\left(14\mathrm{m}\right)}\\ =0.839\mathrm{rad}/{\mathrm{s}}^2\end{array}$

Let $蝇$be angular speed.

Now, according to law of conservation of energy,

$mgh=\frac{1}{2}l{蝇}^2$

Thus $mgh=\left(\frac{1}{2}\right)\left(\frac{1}{3}m{L}^2\right){蝇}^2$

Now $h=\frac{L}{2}\sin 37掳$

Thus the equation becomes $g\left(\frac{L}{2}\right)\sin 37掳=\frac{1}{6}{L}^2{蝇}^2$

Solving for $蝇$,

$\begin{array}{r}蝇=\sqrt{\frac{3g\sin 鈦�{37}^{鈭�}}{L}}\\ =\sqrt{\frac{3\left(9.8\mathrm{m}/{\mathbf{s}}^2\right)\sin 鈦�{37}^{鈭�}}{14\mathrm{m}}}\\ =1.12\mathrm{rad}/{\mathrm{s}}^2\end{array}$

Hence angular speed is $1.12\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$