Two forces, F1 and F2 of magnitude 60 N and 20 N respectively act on a suspended box of weight 30 N as shown in the diagram below. Using the tail-to-head method and a scale of 1 cm represents 10 N, draw an accurate vector diagram to determine the magnitude of the resultant force acting on the box.

The forces acting on the box are:

• $F_1 = 60 \text{ N}$ (upwards)
• $F_2 = 20 \text{ N}$ (upwards)
• Weight $W = 30 \text{ N}$ (downwards)

The scale is $1 \text{ cm}$ represents $10 \text{ N}$.

• Length for $F_1$: $60 \text{ N} \times \frac{1 \text{ cm}}{10 \text{ N}} = 6 \text{ cm}$ (upwards)
• Length for $F_2$: $20 \text{ N} \times \frac{1 \text{ cm}}{10 \text{ N}} = 2 \text{ cm}$ (upwards)
• Length for $W$: $30 \text{ N} \times \frac{1 \text{ cm}}{10 \text{ N}} = 3 \text{ cm}$ (downwards)

1. Draw a vector representing $F_1$, $6 \text{ cm}$ long, pointing upwards.
2. From the head of the $F_1$ vector, draw a vector representing $F_2$, $2 \text{ cm}$ long, also pointing upwards.
3. From the head of the $F_2$ vector, draw a vector representing $W$, $3 \text{ cm}$ long, pointing downwards.

The resultant vector is drawn from the tail of the first vector ($F_1$) to the head of the last vector ($W$).
Since all forces are collinear (acting along the same vertical line), we can find the net displacement:
Total upward length = $6 \text{ cm} + 2 \text{ cm} = 8 \text{ cm}$
Total downward length = $3 \text{ cm}$
Resultant length = Total upward length - Total downward length
Resultant length = $8 \text{ cm} - 3 \text{ cm} = 5 \text{ cm}$ (pointing upwards)

Resultant force magnitude = Resultant length $\times$ Scale factor
Resultant force magnitude = $5 \text{ cm} \times \frac{10 \text{ N}}{1 \text{ cm}} = 50 \text{ N}$

The magnitude of the resultant force is $\text{50 N}$.