In the SI system, scale readings are typically given in kilograms (kg), which represent mass. To calculate torque, first convert this to force (Newtons) using gravity:

$F = m \cdot g$

where $m$ is the scale reading (kg) and $g \approx 9.81 \, \text{m/s}^2$ .

Then calculate torque:

$T = F \cdot a = m \cdot g \cdot a$

Example: A scale reads 10 kg at the end of a 0.3 m wrench. $T = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.3 \, \text{m} = 29.43 \, \text{Nm}$

Imperial Example: Wrench and Luggage Scale

In the Imperial system, scale readings are often in pounds-force (lbf), which is already a unit of force. Torque is calculated directly:

$T = F \cdot a$

Example: A scale reads 20 lbf at the end of a 6-inch wrench. $T = 20 \, \text{lbf} \times 6 \, \text{in} = 120 \, \text{in-lbf}$ Convert to foot-pounds: $120 \, \text{in-lbf} / 12 = 10 \, \text{ft-lbf}$ .

Application: Cylinder Head Torque Setting

A cylinder head requires a torque of 30 Nm. Using a 0.2 m wrench, the required scale reading (mass) can be found by rearranging the torque formula:

$m = \frac{T}{g \cdot a}$

Using $g = 9.81 \, \text{m/s}^2$ : $m = \frac{30 \, \text{Nm}}{9.81 \, \text{m/s}^2 \times 0.2 \, \text{m}} \approx 15.3 \, \text{kg}$

Note: The source material states an approximate reading of 14 kg for this scenario. The discrepancy may stem from using a different value for $g$ (e.g., 10.5 m/s²) or rounding in the diagram.

References

Original Source

Source: engineeringtoolbox.com