You want to loosen a stubbornly tight screw, and have two different handles available for the screwdriver.
If you apply the same force to both handles below, which handle is easier to turn the screw with? In other words, which handle will generate a greater turning moment for the screw?
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Like Tom said, the computation suggests r A F A = r B F B , though on first appearances the definition of torque is a bit opaque.
If r × F doesn't click, we can take a step back and think about things from first principles. Regardless of where the force is applied along the radius, some amount of work must be done to overcome the friction between the screw and whatever it's screwed into.
Let's call this work W unscrew . Then it must be the case that W unscrew is equal to F ( r ) ⋅ path length(r) = F ( r ) ⋅ 2 π r , regardless of the value of r at which we apply the force. Therefore the product r F is a constant.
That's right! If we apply the same force to both screwdrivers, than (A) will generate a larger torque because its distance from the axis is greater.
It is important to take into consideration the correct distance. Having a greater length along the screwdriver won't help as much as having a greater length perpendicular to the axis.
One word... Leverage.... Common sense more than maths for me!
We often have an intuitive feel about how levers work, and how increasing the distance of force from the axis increases the turning moment. We observe this all around us: we open doors by pushing furthest from the hinges. Wheel barrows and nut crackers also work using the same principle.
Also, the 1st one has a handle that is easier to grab
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Using r × F to calculate our torque moment, one can see it is directly proportional to both the length of the moment arm and the applied force.
If the force is the same in both cases, the torque driver A will generate a larger moment as r A > r B .