Why is the output torque less than the input force of 20N applied from a wrench having length of 0.8m?

Question

A wrench is of length 0.8m, or any number below 1.0m, I am exerting a force of 20 Newtons on the point A (one end of the wrench), on point B there is a bolt (on other end).

If I calculate the Torque acting on the bolt, I use the Formula.

Torque = Force X Distance from the Pivot Point.

Therefore

Torque = 20N X 0.8m Torque = 16Nm

1. Can anyone help me understand why the Torque acting on the bolt, less than the input force?
2. Why is there no mechanical advantage? for any value below 1.0
3. If the distance is above 1.0m, I see the Torque is higher than the input force, why? I understand that's how maths multiplication works; I believed that any lever always has some mechanical advantage?

[! Description: Image of an Allen Wrench having length of 0.8m and a force of 20 N being applied through it on a bolt]1

Answer 1

You cannot compare force and torque - they are different attributes measured in different units. Trying to compare $20$ Newtons with $16$ Newton-metres is like trying to compare $20$ seconds with $16$ kilograms.

However, you can compare the forces exerted at each end of the wrench. If the bolt has radius of, say, $1$ cm then the force exerted on the bolt is the torque of $16$ Newton-metres divided by $1$ cm, so it $1600$ Newtons. The mechanical advantage of the wrench is then the ratio of these two forces i.e. $1600$ Newtons divided by $20$ Newtons, which is a mechanical advantage of $80$. This is also the ratio of the two lever arms of the wrench i.e. the ratio of $0.8$ metres to $1$ cm.