A spool of mass m = 360 kg. inner aad ouler radii rho = 1.5 m and R = 2m. respec

Question

A spool of mass m = 360 kg. inner aad ouler radii rho = 1.5 m and R = 2m. respectively and radius of gyration kG = l.8 m, is placed on an incline with theta = 46degree. The cahle that is wrapped around the spool and attached to the wall is initially taut. If the static and kinelic friction coefficients between the incline and the spool are mus = 0.35 and muk = 0.3, respectively. determine the acceleration of G. the angular acceleration of the spool, and the tension in the cable once the spool is released from rest.

Explanation / Answer

First we have to check that the static friction is sufficient to keep the spool or the kinetic friction will act.

Force balance:

Perpendicular to the plane :

N= mgcos(theta) = 2500 N

Max. static friction = 2500*0.35 = 875.3N .............(1)

Along the plane:

T+f=mgSin46 = 2590 N .....................(2)

Torque balance about Centre :

T*r = f*R => 1.5T = 2f =>T=1.33f.....................(3)

Put T in (2)

f = 2590/2.33 = 1112 N which is greater than the friction available

Therefore kinetic friction will act and spool will have motion

So now the force equations becomes:

f = uk * mgCos46 = 750N...............(4)

2590 -T - 750 = 360a

=>T = 1840 - 360a................(5)

Moment equation:

Torque = I*alpha

alpha = a/R

=> 1.5 T - 2*750 = (360*(1.8^2))(a/2)

=> T = 1000 + 388.8 a...............(6)

Solving(5) & (6)

1840 - 360a = 1000 + 388.8 a

=> a = 1.12 m/s^2

T = 1436.15 N

alpha = 0.56 rad/s^2

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