Acceleration of a load.?

Question

A uniform pulley has a mass of 20kg and a radius of 150mm. A rope passes over it and is attached to a load of 10kg. What force must be applied to the other end of the rope to give the load an upward acceleration of 0.2m/s? Can anyone help me out with the formulas for this?

Answer 1

The following assumes a typo in your question and that the acceleration is indeed 2 m/s/s.

The mass of the pulley DOES matter. You not only have to accelerate the 10kg mass, but ALSO the 20kg pulley.

The necessary force will have two components, the first is F = 10kg * (0.2 m/s/s + 9.81 m/s/s). The second F x r = I(pulley) alpha. Where I = the moment of inertia of the pulley and alpha = the angular acceleration of the pulley, which is the tangential acceleration at the outer edge of the pulley (2 m/s/s) divided by the radius of the pulley.

You then add the two to get the total force.

Answer 2

0.2m/s is a velocity, not an acceleration. Something is wrong here.
F=ma, yes, true. But 10kg*0.2m/s is mass*velocity which gives momentum, not force.

Answer 3

sometimes you get little tricks in questions like this - they always seem to occur in 'ball and chain' questions.
mass of pulley - not of interest;
radius of pulley system - nope!

the answer is: F=M*A (mass of load, upward accel~n)
M=10kg, A=0.2m/s, so therefore, F=10*0.2 = 2newtons (N).

hope i'm right? unless i'm getting senile....
g.

Answer 4

From sum of the forces = m * a. We can conclude that the upforce - weight = m * a.

rearranging,

upforce = ma + w = 10*.2 + 10 * 9.8 ((gravity accel)) = 100 N