Question 11.2: The Torque Vector A force of F = (2.00i + 3.00j) N is applie...
The Torque Vector
A force of \overrightarrow{F}=(2.00 \hat{i}+3.00 \hat{j}) N is applied to a rigid object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at a point located at \overrightarrow{r}=(4.00 \hat{i}+5.00 \hat{j}) m relative to the axis. Find the torque \overrightarrow{\tau} applied to the object.
Conceptualize Given the unit-vector notations, think about the directions of the force and position vectors. If this force were applied at this position, in what direction would an object pivoted at the origin turn?
Categorize Because we use the definition of the cross product discussed in this section, we categorize this example as a substitution problem.
Set up the torque vector using Equation 11.1:
\overrightarrow{\tau} \equiv \overrightarrow{r} \times \overrightarrow{F} (11.1)
\overrightarrow{\tau}= \overrightarrow{r} \times \overrightarrow{F}=[(4.00 \hat{i}+5.00 \hat{j}) m] \times[(2.00 \hat{i}+3.00 \hat{j}) N]Perform the multiplication using the distributive law:
\begin{aligned}\overrightarrow{\tau}= & {[(4.00)(2.00) \hat{i} \times \hat{i}+(4.00)(3.00) \hat{i} \times \hat{j}} \\& +(5.00)(2.00) \hat{j} \times \hat{i}+(5.00)(3.00) \hat{j} \times \hat{j}] N \cdot m\end{aligned}Use Equations 11.7a through 11.7d to evaluate the various terms:
\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0 (11.7a)
\hat{i} \times \hat{j}=-\hat{j} \times \hat{i}=\hat{k} (11.7b)
\hat{j} \times \hat{k}=-\hat{k} \times \hat{j}=\hat{i} (11.7c)
\hat{k} \times \hat{i}=-\hat{i} \times \hat{k}=\hat{j} (11.7d)
\overrightarrow{\tau}=[0+12.0 \hat{k}-10.0 \hat{k}+0] N \cdot m=2.0 \hat{k} N \cdot mNotice that both \overrightarrow{r} and \overrightarrow{F} are in the xy plane. As expected, the torque vector is perpendicular to this plane, having only a z component. We have followed the rules for significant figures discussed in Section 1.6, which lead to an answer with two significant figures. We have lost some precision because we ended up subtracting two numbers that are close.