91影视

The weight of the horizontal beam is,190 N .

When a force acts on the component that its fibers are stretched, and its length is increased, then this force is described as the 鈥榯ensile force鈥�.

The tensile force acting on a weighted cable system can be calculated by balancing all the forces acting on the cable in the horizontal and vertical directions.

The free-body diagram can be drawn as,

Here, T is the tension in the cable,$胃$is the angle of the cable from the horizontal beam,$H_v$is vertical, and$H_h$is the horizontal component of the force exerted on the beam at the wall.

The tension in the cable can be divided into two components, the vertical component T , $\sin 胃$and the horizontal component Tcos$胃$.

From a right-angled triangle, we can write,

$\begin{array}{r}\sin 鈦�胃=\frac{3}{5}\\ \cos 鈦�胃=\frac{4}{5}\end{array}$

Taking moments about point O, we get,

$\begin{array}{r}T\sin 鈦�胃脳4\mathrm{m}鈭�300\mathrm{N}脳4\mathrm{m}鈭�190\mathrm{N}脳2\mathrm{m}=0\\ 4T\sin 鈦�胃\mathrm{m}鈭�1200\mathrm{N}鈰�\mathrm{m}鈭�380\mathrm{N}鈰�\mathrm{m}=0\\ 4T\sin 鈦�胃\mathrm{m}=1580\mathrm{N}鈰�\mathrm{m}\\ T=\frac{1580\mathrm{N}鈰�\mathrm{m}}{4\sin 鈦�胃\mathrm{m}}\end{array}$

Putting the values in the above expression, we get,

$\begin{array}{r}T=\frac{1580\mathrm{N}鈰�\mathrm{m}}{4脳\frac{3}{5}\mathrm{m}}\\ T=\frac{1580脳5}{12}\mathrm{N}\\ T=658.33\mathrm{N}\end{array}$

Hence, the tension in the cable is 658.33 N .

Balancing all the forces in the vertical direction, and we get,

$\begin{array}{r}T\sin 鈦�胃+H_v=190\mathrm{N}+300\mathrm{N}\\ T\sin 鈦�胃+H_v=490\mathrm{N}\\ H_v=490\mathrm{N}鈭�T\sin 鈦�胃\end{array}$

Putting the values in the above expression and we get,

$\begin{array}{r}{H}_v=490\mathrm{N}鈭�658.33\mathrm{N}脳\frac{3}{5}\\ H_v=490\mathrm{N}鈭�658.33\mathrm{N}脳\frac{3}{5}\\ H_v=490\mathrm{N}鈭�395\mathrm{N}\\ H_v=95\mathrm{N}\end{array}$

Balancing all the forces in the horizontal direction, and we get,

$\begin{array}{r}{H}_h鈭�T\cos 鈦�胃=0\\ H_h=T\cos 鈦�胃\end{array}$

Putting the values in the above expression and we get,

$\begin{array}{r}{H}_h=658.33\mathrm{N}脳\frac{4}{5}\\ H_h=526.664\mathrm{N}\end{array}$

Hence, the horizontal and vertical components of the force exerted on the beam at the wall are 526.664 N and 95 N respectively.