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A force $\mathbf{F}_1=A \hat{\mathbf{j}}$ is applied to a point, whose radius vector $\mathbf{r}_1=a \hat{\mathbf{i}}$, while a force $\mathbf{F}_2=B \hat{\mathbf{i}}$ is applied to the point whose radius vector $\mathbf{r}_2=\hat{b} \hat{\mathbf{j}}$. Both the radius vectors are determined relative to the origin of the coordinate axes $O$. The moment of the force relative to $O$ is
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Verified Answer
The correct answer is:
$(a A-b B) \hat{\mathbf{k}}$
Given that,
$$
\begin{aligned}
& \mathbf{F}_1=A \hat{\mathbf{j}}, \mathbf{r}_1=a \hat{\mathbf{i}} \\
& \mathbf{F}_2=B \hat{\mathbf{i}}, \mathbf{r}_2=b \hat{\mathbf{j}}
\end{aligned}
$$
First Case
$$
\begin{aligned}
& \tau_1=\text { moment due to } \mathbf{F}_1=\mathbf{r}_1 \times \mathbf{F}_1 \\
& \Rightarrow \quad \tau_1=a \hat{\mathbf{i}} \times A \hat{\mathbf{j}} \\
& \tau_1=a A \hat{\mathbf{k}} \\
&
\end{aligned}
$$
Second case
$$
\begin{aligned}
\tau_2 & =\mathbf{r}_2 \times \mathbf{F}_2 \\
\tau_2 & =\hat{b} \times \hat{\mathbf{j}} \times \hat{B} \\
\tau_2 & =-b \hat{\mathbf{k}}
\end{aligned}
$$
So, net moment is $\tau=\tau_1+\tau_2$
$$
\begin{aligned}
\Rightarrow \quad \tau & =(a A+(-b B) \hat{\mathbf{k}} \\
\tau & =(a A-b B) \hat{\mathbf{k}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathbf{F}_1=A \hat{\mathbf{j}}, \mathbf{r}_1=a \hat{\mathbf{i}} \\
& \mathbf{F}_2=B \hat{\mathbf{i}}, \mathbf{r}_2=b \hat{\mathbf{j}}
\end{aligned}
$$
First Case
$$
\begin{aligned}
& \tau_1=\text { moment due to } \mathbf{F}_1=\mathbf{r}_1 \times \mathbf{F}_1 \\
& \Rightarrow \quad \tau_1=a \hat{\mathbf{i}} \times A \hat{\mathbf{j}} \\
& \tau_1=a A \hat{\mathbf{k}} \\
&
\end{aligned}
$$
Second case
$$
\begin{aligned}
\tau_2 & =\mathbf{r}_2 \times \mathbf{F}_2 \\
\tau_2 & =\hat{b} \times \hat{\mathbf{j}} \times \hat{B} \\
\tau_2 & =-b \hat{\mathbf{k}}
\end{aligned}
$$
So, net moment is $\tau=\tau_1+\tau_2$
$$
\begin{aligned}
\Rightarrow \quad \tau & =(a A+(-b B) \hat{\mathbf{k}} \\
\tau & =(a A-b B) \hat{\mathbf{k}}
\end{aligned}
$$
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