Question

Find the torque about the origin when a force of $3 \,\hat{j}\, N$ acts on a particle whose position vector is $2\, \hat {k}\, m$ .

• A. $6 \hat{i}\, Nm $
• B. $6 \hat{j} \,Nm $
• C. $-6 \hat{i} \,Nm $
• D. $6 \hat{k} Nm $

Answer 1

The Correct Option is C

To find the torque about the origin when a force acts on a particle, we utilize the formula for torque:

\(\mathbf{\tau} = \mathbf{r} \times \mathbf{F}\)

where:

• \(\mathbf{\tau}\) is the torque.
• \(\mathbf{r}\) is the position vector of the particle.
• \(\mathbf{F}\) is the force vector.

In this problem, the force vector is given by \(\mathbf{F} = 3 \,\hat{j}\, N\), and the position vector is given by \(\mathbf{r} = 2 \,\hat{k}\, m\).

Let's calculate the cross product \(\mathbf{r} \times \mathbf{F}\):

\(\mathbf{\tau} = (2 \,\hat{k}) \times (3 \,\hat{j})\)

According to the rules of the cross product for unit vectors:

• \(\hat{i} \times \hat{j} = \hat{k}\)
• \(\hat{j} \times \hat{k} = \hat{i}\)
• \(\hat{k} \times \hat{i} = \hat{j}\)
• \(\hat{j} \times \hat{i} = -\hat{k}\)
• \(\hat{k} \times \hat{j} = -\hat{i}\)
• \(\hat{i} \times \hat{k} = -\hat{j}\)

Using \(\hat{k} \times \hat{j} = -\hat{i}\), we have:

\((2 \hat{k}) \times (3 \hat{j}) = 2 \times 3 \,(-\hat{i}) = -6 \hat{i}\)

Therefore, the torque about the origin is \(-6 \hat{i} \,Nm\).

Thus, the correct answer is:

\(-6 \hat{i} \,Nm\)

This matches the correct answer option provided in the question.