Animated Solution for Physics - Rotational Motion: A slab is subjected to two forces ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$ of same magnitude $F$ as shown in the figure. Force ${\mathbf{F}}_2$ is in $xy$-plane while force ${\mathbf{F}}_1$ acts along $Z$-axis at the point $(2\hat{\mathbf{i}}+3\hat{\mathbf{j}})$. The moment of these forces about point $O$ will be

## Mastering 3D Torque: A Vector Approach

Imagine you are trying to twist a heavy rectangular slab fixed at its corner. The twisting effect, or torque, depends not just on how hard you push, but exactly where and in what direction you apply the force. This problem is a beautiful exercise in 3D vector mechanics, requiring us to carefully track position vectors and execute cross products without losing our way in the coordinate system.

Let's break down the physical situation. We have a slab lying in the $xy$-plane. Two forces, ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$, both of magnitude $F$, are acting on it. To find the total moment (torque) about the origin $O$, we must first identify the position vectors for the points of application of these forces.

Force ${\mathbf{F}}_1$ acts at the point $(2,3,0)$. Therefore, its position vector is:

Force ${\mathbf{F}}_2$ acts on the $y$-axis at a distance of $6\text{ m}$ from the origin. Its position vector is simply:

The fundamental definition of torque about a point is the cross product of the position vector and the force vector:

Since torque is a linear operator, the net torque on the slab is the vector sum of the individual torques:

Force ${\mathbf{F}}_1$ acts vertically upwards, parallel to the $Z$-axis. We can express it as:

Now, we set up the cross product for the first torque:

Distributing the cross product, we must strictly follow the cyclic order of unit vectors ($\hat{\mathbf{i}}\times \hat{\mathbf{k}}=-\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}\times \hat{\mathbf{k}}=\hat{\mathbf{i}}$):

Force ${\mathbf{F}}_2$ is a bit trickier. It lies entirely in the $xy$-plane and makes an angle of $30^{\circ }$ with the negative $y$-axis, pointing into the third quadrant. Resolving it into its Cartesian components gives:

Now, we calculate the second torque:

When we distribute this, the $\hat{\mathbf{j}}\times \hat{\mathbf{j}}$ term vanishes because the cross product of any vector with itself is zero. We are left with:

Since $\hat{\mathbf{j}}\times \hat{\mathbf{i}}=-\hat{\mathbf{k}}$ and $\sin 30^{\circ }=\frac{1}{2}$, we get:

To find the grand total, we simply add our two torque vectors together:

Factoring out the magnitude $F$, we arrive at our final, elegant result:

This perfectly matches option (b). The beauty of the vector cross product is that it automatically handles all the complex 3D geometry for us, provided we are meticulous with our signs and unit vectors!