Calculating Torque Required for Angular Motion of a Winding Drum

[Bikerz]

Question is

A problem has arisen in teh goods yard where a large container is required to be pulled along a makeshift alipway. The container has a mass of 1.5tonne and is pulled along the horizontal slipway at an acceleration of 0.9m/s² by a rope wound around a winding drum. The resistance to motion of the container is equivalent to a coefficient of friction 0.6. The winidng drum hasa mass of 0.5 tonne, outside diameter 1m and radius of gyration of 0.3m. The drum has a resistance couple of 200Nm acting on it. Find teh torque required by the motor of teh winidng drum?

I have so far:



Q(Anglaur acceleration, can't get right sybol)


I=mk²
500x0.3² =45kgm²


Q=a/r

Q=0.9/0.5
=1.8 rad/s²

f=ma
p-f=ma
p=ma + f

(1500x0.9) + 0.6
=1350.6N


∑T=IK
E-Pr-R=IK
E=IQ+Pr+R
(45x1.8)+(1350x0.6x0.5) +200
=956.3Nm

That correct?

Cheers

 

[nvn]

Bikerz: Wouldn't the friction force be F = 0.6*m*g, not 0.6? Try it again.

 

[Bikerz]

Ah yes cheers. 810N

Thanks

Im rubbish at taking notes, scribble down as fast as I can go to keep up.