A bar of length 3m is subjected to two parallel but opposite forces of 60N at its ends as shown in the diagram above. The couple on the bar is

Step 1: Understand the concept of a couple. A couple is a pair of forces of equal magnitude, acting in opposite directions, and whose lines of action are parallel but do not coincide. The effect of a couple is to produce rotation (torque) without causing any translational motion. The magnitude of the couple (or torque) is given by the product of the magnitude of one of the forces and the perpendicular distance between their lines of action.

Step 2: Identify the given values from the problem and diagram. • Magnitude of each force, $F = 60 \text{ N}$. • Length of the bar (distance between the points of application of the forces), $L = 3 \text{ m}$. • Angle between the forces and the bar, $\theta = 30^\circ$.

Step 3: Determine the perpendicular distance between the lines of action of the forces. The perpendicular distance ($d_\perp$) between the lines of action of the two parallel forces can be found using trigonometry. If the forces are applied at the ends of a bar of length $L$ and make an angle $\theta$ with the bar, then the perpendicular distance between their lines of action is $d_\perp = L \sin\theta$.

Step 4: Calculate the perpendicular distance. $d_\perp = L \sin\theta$ $d_\perp = 3 m \times \sin(30^\circ)$ We know that $\sin(30^\circ) = \frac{1}{2}$. $d_\perp = 3 m \times \frac{1}{2}$ $d_\perp = 1.5 \text{ m}$

Step 5: Calculate the magnitude of the couple. The magnitude of the couple $C$ is given by: $C = F \times d_\perp$ Substitute the values of $F$ and $d_\perp$: $C = 60 N \times 1.5 \text{ m}$ $C = 90 \text{ Nm}$

The final answer is $\boxed{90 Nm}$.