Rotation Ques 9
9 A slab is subjected to two forces $\mathbf{F}_1$ and $\mathbf{F}_2$ of same magnitude $F$ as shown in the figure. Force $\mathbf{F}_2$ is in $x y$-plane while force $\mathbf{F}_1$ acts along $Z$-axis at the point $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ ). The moment of these forces about point $O$ will be
(2019 Main, 11 Jan I)
(a) $(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) F$
(b) $(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) F$
(c) $(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) F$
(d) $(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) F$
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Answer:
Correct Answer: 9.( b )
Solution:
$ \begin{aligned} & \mathbf{r}_1=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \text { and } \mathbf{r}_2=6 \hat{\mathbf{j}} \\ & \mathbf{F}_1=F \hat{\mathbf{k}} \\ & \text { and } \quad \mathbf{F}_2=\left(-\sin 30^{\circ} \hat{\mathbf{i}}-\cos 30^{\circ} \hat{\mathbf{j}}\right) \mathrm{F} \\ & \end{aligned} $
Moment of force is given as, $\tau=\mathbf{r} \times \mathbf{F}$ where, $\mathbf{r}$ is the perpendicular distance and $\mathbf{F}$ is the force.
$\therefore$ Moment due to $\mathbf{F}_1$
$ \begin{aligned} \tau_1 & =(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \times(F \hat{\mathbf{k}}) \\ & =-2 F \hat{\mathbf{j}}+3 F \hat{\mathbf{i}} \end{aligned} $
Moment due to $\mathbf{F}_2$
$ \begin{aligned} \tau_2 & =(6 \hat{\mathbf{j}}) \times\left(-\sin 30^{\circ} \hat{\mathbf{i}}-\cos 30 \hat{\mathbf{j}}\right) F \\ & =6 \sin 30^{\circ} \mathrm{F} \hat{\mathbf{k}}=3 \mathrm{~F} \hat{\mathbf{k}} \end{aligned} $
$\therefore$ Resultant torque,
$ \begin{aligned} \tau & =\tau_1+\tau_2=3 F \hat{\mathbf{i}}-2 F \hat{\mathbf{j}}+3 F \hat{\mathbf{k}} \\ & =(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) F . \end{aligned} $
BETA
