Question

Find the torque of a force $ \mathbf{\vec{F}}=-3\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}} $ acting at the point $ \vec{r}=-7\mathbf{\hat{i}}+3\mathbf{\hat{j}}+16\mathbf{\hat{k}} $ :

• A. $ -21\mathbf{\hat{i}}+3\mathbf{\hat{j}}+5\mathbf{\hat{k}} $
• B. $ -14\mathbf{\hat{i}}+3\mathbf{\hat{j}}+16\mathbf{\hat{k}} $
• C. $ 4\mathbf{\hat{i}}+4\mathbf{\hat{j}}+6\mathbf{\hat{k}} $
• D. $ 14\mathbf{\hat{i}}+38\mathbf{\hat{j}}+16\mathbf{\hat{k}} $

Answer 1

Answer: (D)

Key Idea: The torque of a force is the cross product of $ \overrightarrow{\mathbf{r}} $ and $ \overrightarrow{\mathbf{F}} $ in the same order. Given: $ \overrightarrow{\mathbf{r}}=7\widehat{\mathbf{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}},\,\overrightarrow{\mathbf{F}}=-3\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}} $ $ \overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\mathbf{\hat{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}})\,\times (-3\mathbf{\hat{i}}+\mathbf{j}+5\mathbf{\hat{k}}) $ $ =\left| \begin{matrix} {\mathbf{\hat{i}}} & {\mathbf{\hat{j}}} & {\mathbf{\hat{k}}} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \\ \end{matrix} \right| $ $ =\mathbf{\hat{i}}(15-1)\mathbf{\hat{j}}\,(35+3)+\mathbf{\hat{k}}(7+9) $ $ =14\mathbf{\hat{i}}-38\mathbf{\hat{j}}+16\mathbf{\hat{k}} $ Alternative: $ \overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\mathbf{\hat{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}})\times (-3\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}}) $ $ =-21(\mathbf{\hat{i}}\times \mathbf{\hat{i}})+7(\mathbf{\hat{i}}\times \mathbf{\hat{j}})+35(\mathbf{\hat{i}}\times \mathbf{\hat{k}}) $ $ -9(\mathbf{\hat{j}}\times \mathbf{\hat{i}})+3(\mathbf{\hat{j}}\times \mathbf{\hat{j}})+15(\mathbf{\hat{j}}\times \mathbf{\hat{k}}) $ $ -3(\mathbf{\hat{k}}\times \mathbf{\hat{i}})+(\mathbf{\hat{k}}\times \mathbf{\hat{j}})+5(\mathbf{\hat{k}}\times \mathbf{\hat{k}}) $ $ =0+7\mathbf{\hat{k}}-35\mathbf{\hat{j}}+9\mathbf{\hat{k}}+0+15\mathbf{\hat{i}}-3\mathbf{\hat{j}}-\mathbf{\hat{i}}=0 $ $ =14\mathbf{\hat{i}}-38\mathbf{\hat{j}}+16\mathbf{\hat{k}} $