Question

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

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

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: $\overrightarrow{\mathbf{r}}=7\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}},\overrightarrow{\mathbf{F}}=-3\hat{\mathbf{i}}+\hat{\mathbf{j}}+5\hat{\mathbf{k}}$ $\overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}})\times (-3\hat{\mathbf{i}}+\mathbf{j}+5\hat{\mathbf{k}})$ $=\left|\begin{array}{ccc}\hat{\mathbf{i}}& \hat{\mathbf{j}}& \hat{\mathbf{k}}\\ 7& 3& 1\\ -3& 1& 5\end{array}\right|$ $=\hat{\mathbf{i}}(15-1)\hat{\mathbf{j}}(35+3)+\hat{\mathbf{k}}(7+9)$ $=14\hat{\mathbf{i}}-38\hat{\mathbf{j}}+16\hat{\mathbf{k}}$ Alternative: $\overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}})\times (-3\hat{\mathbf{i}}+\hat{\mathbf{j}}+5\hat{\mathbf{k}})$ $=-21(\hat{\mathbf{i}}\times \hat{\mathbf{i}})+7(\hat{\mathbf{i}}\times \hat{\mathbf{j}})+35(\hat{\mathbf{i}}\times \hat{\mathbf{k}})$ $-9(\hat{\mathbf{j}}\times \hat{\mathbf{i}})+3(\hat{\mathbf{j}}\times \hat{\mathbf{j}})+15(\hat{\mathbf{j}}\times \hat{\mathbf{k}})$ $-3(\hat{\mathbf{k}}\times \hat{\mathbf{i}})+(\hat{\mathbf{k}}\times \hat{\mathbf{j}})+5(\hat{\mathbf{k}}\times \hat{\mathbf{k}})$ $=0+7\hat{\mathbf{k}}-35\hat{\mathbf{j}}+9\hat{\mathbf{k}}+0+15\hat{\mathbf{i}}-3\hat{\mathbf{j}}-\hat{\mathbf{i}}=0$ $=14\hat{\mathbf{i}}-38\hat{\mathbf{j}}+16\hat{\mathbf{k}}$