Figure 10-25bshows an overhead view of a horizontal bar that is rotated about the pivot point by two horizontal forces${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}$, and ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$with${\overrightarrow{\mathbf{F}}}_{\mathbf{3}}$at angle$\mathit{\varphi }$to the bar. Rank the following values of $\mathit{\varphi }$according to the magnitude of the angular acceleration of the bar, greatest first:$\mathbf{90}\mathbf{°}\mathbf{,}\mathbf{}\mathbf{70}\mathbf{°}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{110}\mathbf{°}\mathbf{.}$

Angles$70°,\hspace{0.33em}90°\mathrm{and}110°$

Torque is equal to the moment of force. It can also be written as the product of moment of inertia and angular acceleration.

Find the torque related to each angle. Then, using the relation between torque and angular acceleration rank the angles according to the magnitude of angular acceleration.

Formulae are as follows:

$$\begin{gathered} \tau =la \\ =r\times F \\ =rF\sin \varphi \end{gathered}$$

Where, r is radius, F is force and τ is torque.

Here, simultaneously calculate torque for each case.

$\tau =rFsin\varphi ,$, as torque only depends on angle. So the greater the angle greater will be torque.

$$\begin{gathered} 1.sin\varphi =sin90=1 \\ 2.sin\varphi =sin70=0.9396 \\ 3.sin\varphi =sin110=0.9396 \end{gathered}$$

Therefore,${\tau }_1>{\tau }_2={\tau }_3$

Since,$\tau =I\alpha$ ,

${\alpha }_1>{\alpha }_2={\alpha }_3$

The rank of the angle according to the magnitude of angular acceleration is,$90°>70°=110°$

Therefore, using the given diagram of forces and values of angles and proportionality between torque and angular acceleration the forces can be ranked.