Abucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. a crank with a turning radius of 0.25 m is attached to the end of the cylinder. what minimum force directed perpendicular to the crank handle is required to just raise the buc

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

   = 11.27 Nm

Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

  = 45.08 N

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