Extending a winch handle

[sparkythecat]

I have a winch with a handle that is 300mm from the axis of the drum.
If I extend the length by 75mm, so that the distance between the handle and the axis of the drum is now 375mm, by what percentage, will I decrease the force necessary to turn the drum under the same load?

Is the answer 25%?

[V8LM]

Torque = force * distance.

To gain the same torque, if you increase the distance by 1.25, you decrease the force to 80% of its original value (1/1.25).

[herewego]

Depends on the radius of the load cable, i.e. drum plus cable. The force needed on the handle will be pulling load x drum rad / 300 etc.

[sparkythecat]

Can you just expand on that a bit?
The radius of the drum at the start of the lift is 40mm and as the cable is wound onto the drum the radius will increase to 70mm at the end of the lift.

[bearman68]

V8LM said:

Torque = force * distance.

To gain the same torque, if you increase the distance by 1.25, you decrease the force to 80% of its original value (1/1.25).

This

[herewego]

sparkythecat said:

Can you just expand on that a bit?
The radius of the drum at the start of the lift is 40mm and as the cable is wound onto the drum the radius will increase to 70mm at the end of the lift.

You could just use an average of 55. Obviously greater force will be required on the handle as the drum fills up.

[V8LM]

sparkythecat said:

Can you just expand on that a bit?
The radius of the drum at the start of the lift is 40mm and as the cable is wound onto the drum the radius will increase to 70mm at the end of the lift.

The force required when the lift is at the top is 1.75 times that required when the lift is at the bottom (actually slightly less because when at the bottom the weight of the lift includes that of the cable). But to your original question about the change in force required on extending the lever - it's still 80%


When the lift is at the bottom the torque produced by the lift on the drum is

weight * 40

and when the lift is at the top the torque is

weight * 70


With your original 300 mm lever, when the lift is at the bottom then the force you need to apply to counteract the torque is

weight * 40 / 300

and when at the top it's

weight * 70 / 300


Extend the lever to 375, and the force when at the bottom is

weight * 40 / 375

and at the top is

weight * 70 / 375


The fractional change in force is therefore

(weight * 40 / 375) / (weight * 40 / 300) = 300 / 375 = 0.8

and

(weight * 70 / 375) / (weight * 70 / 300) = 300 / 375 = 0.8


(where 'weight' can include that the cable and is the mass * g)

Edited by V8LM on Saturday 30th April 09:21