91影视

The condition for translational equilibrium is: $鈭�F_{ext}=0$.

And that for rotational equilibrium is: $鈭�{蟿}_{ext}=0$.

Given, the strut and the crate, both weigh w Newton. Let the length of the strut be L units.

Now, in part (a), to set up euilibrium, sum of all forces meant to be zero irrespective of directions. So, taking forces of horizontal direction, we get:

$$\begin{gathered} 鈭�F_x-T=0 \\ {F}_x=T鈥�鈥� \end{gathered}$$ ...........(i)

Similarly, taking all forces in the vertical direction, we have:

$$\begin{gathered} 鈭�F_y-w-w=0 \\ {F}_y=2w \end{gathered}$$ ...............(ii)

Also, the equation for the torque $蟿$in the given system will be:

$$\begin{gathered} 鈭�蟿=0 \\ -w路\left(\frac{L}{2}路\cos 30掳\right)-w路\left(L路\cos 30掳\right)+T路\left(L路\sin 30掳\right)=0 \\ T路\left(L路\sin 30掳\right)=w路\left(\frac{3L}{2}路\cos 30掳\right) \\ T路\sin 30{}^{鈭�}=\frac{3w}{2}路\cos 30{}^{鈭�} \\ T=\frac{3}{2路\tan 30掳}路w \\ T=2.598w鈥� \end{gathered}$$ .............(iii)

Now, from equations (i) and (iii), we have:

$$\begin{gathered} F_x=T \\ =2.598w \end{gathered}$$ ................(iv)

Also, from equations (ii) and (iv), the magnitude of the resultant force will be:

$$\begin{gathered} F=\sqrt{F_x^2+F_y^2} \\ =\sqrt{{\left(2.598w\right)}^2+{\left(2w\right)}^2} \\ =3.279w \end{gathered}$$

And the direction of the resultant force will be given by:

$$\begin{gathered} \tan 胃=\frac{F_x^{}}{F_y^{}} \\ \tan 胃=\frac{2.598}{2} \\ 胃={\tan }^{-1}\left(\frac{2.598}{2}\right) \\ 胃=52.41掳 \end{gathered}$$

Hence, the required value in part (a) are:

$$\begin{gathered} T=2.60w \\ F=3.28w \\ 胃=52.41掳 \end{gathered}$$

Now, in part (b), unlike part (a), the Tension here will now have the two components, that are $T\sin 30{}^{鈭�}\text{and}T\cos 30掳$. So, taking forces of horizontal, we have:

$$\begin{gathered} 鈭�F_x-T\cos 30{}^{鈭�}=0 \\ {F}_x=T\cos 30{}^{鈭�}鈥� \end{gathered}$$ ...............(v)

Similarly, taking all forces in the vertical direction, we have:

$$\begin{gathered} 鈭�F_y-w-w-T\sin 30{}^{鈭�}=0 \\ {F}_y=2w+T\sin 30{}^{鈭�}鈥� \end{gathered}$$ .................(vi)

Also, the equation for the torque $蟿$in the given system will be:

$$\begin{gathered} 鈭�蟿=0 \\ -w路\left(\frac{L}{2}路\cos 45掳\right)-w路\left(L路\cos 45掳\right)+T\sin 30{}^{鈭�}路\left(L路\cos 45掳\right)-T\cos 30{}^{鈭�}路\left(L路\cos 45掳\right)=0 \\ \frac{3w}{2}+T路\left(\sin 30{}^{鈭�}-\cos 30掳\right)=0 \\ \frac{3w}{2}+T路\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)=0 \\ T=4.098w鈥� \end{gathered}$$ ...............(vii)

Now, from equations (v) and (vii), we have:

$$\begin{gathered} F_x=T\cos 30{}^{鈭�} \\ =0.866T \\ =3.549w鈥� \end{gathered}$$ ..............(viii)

And,

$$\begin{gathered} F_y=2w+T\sin 30{}^{鈭�} \\ =2w+0.5T \\ =4.049w鈥� \end{gathered}$$.................(ix)

Also, from equations (viii) and (ix), the magnitude of the resultant force will be:

$$\begin{gathered} F=\sqrt{F_x^2+F_y^2} \\ =\sqrt{{\left(3.549w\right)}^2+{\left(4.049w\right)}^2} \\ =5.384w \end{gathered}$$

And the direction of the resultant force will be given by:

$$\begin{gathered} \tan 胃=\frac{F_x^{}}{F_y^{}} \\ \tan 胃=\frac{4.049}{3.549} \\ 胃={\tan }^{-1}\left(\frac{4.049}{3.549}\right) \\ 胃=48.765掳 \end{gathered}$$

Hence, the required values in part (b) are:

$$\begin{gathered} T=4.10w \\ F=5.38w \\ 胃=48.8掳 \end{gathered}$$