Calculate the net moment about pivot 'o' given two forces: 50 N at 3 m to the left and 80 N at 4 m to the right.

The principle of moments states that the moment (or torque) about a pivot is the product of the force and its perpendicular distance from the pivot. Moments causing anticlockwise rotation are typically taken as positive, and moments causing clockwise rotation as negative.

• Force $F_1 = 50 \text{ N}$ acts at a distance $d_1 = 3 \text{ m}$ to the left of the pivot. This force creates an anticlockwise moment.
• Force $F_2 = 80 \text{ N}$ acts at a distance $d_2 = 4 \text{ m}$ to the right of the pivot. This force creates a clockwise moment.

The moment due to the $50 \text{ N}$ force is:

$$\tau_{anticlockwise} = F_1 \times d_1$$ $$\tau_{anticlockwise} = 50 \text{ N} \times 3 \text{ m}$$ $$\tau_{anticlockwise} = 150 \text{ Nm}$$

The moment due to the $80 \text{ N}$ force is:

$$\tau_{clockwise} = F_2 \times d_2$$ $$\tau_{clockwise} = 80 \text{ N} \times 4 \text{ m}$$ $$\tau_{clockwise} = 320 \text{ Nm}$$

The net moment is the sum of the moments, considering their directions. Taking anticlockwise moments as positive and clockwise moments as negative:

$$\tau_{net} = \tau_{anticlockwise} - \tau_{clockwise}$$ $$\tau_{net} = 150 \text{ Nm} - 320 \text{ Nm}$$ $$\tau_{net} = -170 \text{ Nm}$$

The negative sign indicates that the net moment is in the clockwise direction.

The net moment about point 'o' is $\text{170 Nm clockwise}$.