91影视

When the rates of the forward and reverse reactions are equal, a system is said to be in equilibrium. The pace of the forward reaction rises with the addition of more reactants. It appears that the equilibrium is shifting toward the product, or right, side of the equation since the rate of the reverse reaction remains unchanged at first.

Here we have, the length of the uniform steel rod is

Tension T = 650 N

The rod is to be held at $22.0掳$below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. So,${胃}_1=22.0掳$

The cable makes an angle of $30.0掳$with the rod and is attached to the wall at a point above the hinge${胃}_2=30.0掳$ .

(a)

The rod is at the equilibrium state, so the summation of the torque at any point of the rod is zero

So

$\begin{array}{r}鈭�蟿=0\\ \left(鈭�T\sin 鈦�{30}^{鈭�}\right)\left(8m\right)+\left(mg\cos 鈦�{22}^{鈭�}\right)\left(4m\right)=0\end{array}$

$\begin{array}{r}m=\frac{\left(T\sin 鈦�{30}^{鈭�}\right)\left(8m\right)}{\left(g\cos 鈦�{22}^{鈭�}\right)\left(4m\right)}\\ m=\frac{\left(650N\right)\left(\sin 鈦�{30}^{鈭�}\right)\left(8m\right)}{\left(9.8m/s^2\right)\left(\cos 鈦�{22}^{鈭�}\right)\left(4m\right)}\\ m=71.5\mathrm{kg}\end{array}$

Hence the mass of the rod is 71.5 kg .

(b)

Let the rod has a mass that isless than the value calculated in part (a)

So, in this case, the mass is

m = 71.5 kg - 10.0 kg

= 61.5 kg .

Now, from equation (1)

$\begin{array}{r}\mathrm{T}=\frac{\left(\mathrm{mgcos}鈦�{22}^{鈭�}\right)\left(4\mathrm{m}\right)}{\left(\sin 鈦�{30}^{鈭�}\right)\left(8\mathrm{m}\right)}\\ \mathrm{T}=\frac{\left(61.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\cos 鈦�{22}^{鈭�}\right)\left(4\mathrm{m}\right)}{\left(\sin 鈦�{30}^{鈭�}\right)\left(8\mathrm{m}\right)}\\ \mathrm{T}=559\mathrm{N}\end{array}$

Now, the hinge force is applied with two components${\mathrm{F}}_{\mathrm{x}}$and${\mathrm{F}}_{\mathrm{y}}$.

The vertical force of the hinge is in opposite direction to the tension and in the same direction of the weight,

Therefore we have,

$\begin{array}{r}T\cos 鈦�{38}^{鈭�}鈭�mg鈭�F_y=0\\ F_y=T\cos 鈦�{38}^{鈭�}鈭�mg\end{array}$

Put the value of T , m and g in the above equation,

$\begin{array}{r}{F}_y=T\cos 鈦�{38}^{鈭�}鈭�mg\\ F_y=\left(559\mathrm{N}\right)\cos 鈦�{38}^{鈭�}鈭�\left(61.5\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\\ F_y=鈭�162.2\mathrm{N}\end{array}$

The horizontal force$F_x$of the hinge is in opposite direction to T sin$38掳$.

Therefore we have,

$\begin{array}{r}\mathrm{Tsin}鈦�{38}^{鈭�}鈭�{\mathrm{F}}_{\mathrm{x}}=0\\ {\mathrm{F}}_{\mathrm{x}}=\mathrm{Tsin}鈦�{38}^{鈭�}\\ {\mathrm{F}}_{\mathrm{x}}=\left(559\mathrm{N}\right)\sin 鈦�{38}^{鈭�}\\ {\mathrm{F}}_{\mathrm{x}}=344\mathrm{N}\\ \mathrm{Now},\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{force}\mathrm{of}\mathrm{the}\mathrm{hinge}\mathrm{is}\\ \begin{array}{rl}{\mathrm{F}}_{\mathrm{h}}& =\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2}\\ & =\sqrt{(344\mathrm{N}{)}^2+(鈭�162.2\mathrm{N}{)}^2}\\ & =380\mathrm{N}\end{array}\end{array}$

To find the angle of direction of ${\mathrm{F}}_{\mathrm{h}}$,

$\begin{array}{r}\tan 鈦�胃=\frac{F_y}{F_x}\\ 胃={\tan }^{鈭�1}鈦�\left(\frac{F_y}{F_x}\right)\\ 胃={\tan }^{鈭�1}鈦�\left(\frac{162.2\mathrm{N}}{344\mathrm{N}}\right)\\ 胃={25.2}^{鈭�}\end{array}$

The magnitude of the force that the hinge exerts on the rod is 380 N

The direction of the force that the hinge exerts on the rod is $25.2掳$.