To solve for the force [tex]\( F \)[/tex] that produces a torque (or moment) of 10.5 Nm (Newton-meters) using a spanner, we need to understand the relationship between force, torque, and distance.
### Step-by-Step Solution
1. Identify the known values:
- The perpendicular distance [tex]\( d \)[/tex] between the axis of the nut and the line of action of the force applied is [tex]\( 0.15 \)[/tex] meters.
- The torque [tex]\( \tau \)[/tex] generated by the force is [tex]\( 10.5 \)[/tex] Nm.
2. Understand the relationship between force, torque, and distance:
- Torque ([tex]\( \tau \)[/tex]) is given by the product of the force ([tex]\( F \)[/tex]) and the perpendicular distance ([tex]\( d \)[/tex]):
[tex]\[
\tau = F \times d
\][/tex]
- Here, the force is perpendicular to the axis, which directly applies to this formula.
3. Rearrange the formula to solve for the force [tex]\( F \)[/tex]:
- We need to isolate [tex]\( F \)[/tex] in the equation:
[tex]\[
F = \frac{\tau}{d}
\][/tex]
4. Substitute the given values into the equation:
- Substitute [tex]\( \tau = 10.5 \, \text{Nm} \)[/tex] and [tex]\( d = 0.15 \, \text{m} \)[/tex]:
[tex]\[
F = \frac{10.5}{0.15}
\][/tex]
5. Calculate the force [tex]\( F \)[/tex]:
- Perform the division:
[tex]\[
F = 70 \, \text{N}
\][/tex]
### Conclusion
The magnitude of the force [tex]\( F \)[/tex] needed to produce a torque of 10.5 Nm using a spanner with a perpendicular distance of 0.15 meters is [tex]\(\boxed{70 \, \text{N}}\)[/tex].
