A plank with a mass M = 6.60 kg rides on top of two identical, solid, cylindrica

Question

A plank with a mass M = 6.60 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.90 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force of magnitude 5.40 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

(a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank.

(b) Find the acceleration of the rollers at this moment.

(c) What friction forces are acting at this moment? (Let fp be the frictional force exerted by each roller on the plank, and let fg be the rolling friction exerted by the ground on each roller.)

Explanation / Answer

let the acceleration of plank be a1 and that of roller is a2
and angular acceleration be

since rollers are under pure rolling in a clockwise direction

a2=*r

applying point of contact equation between plank and roller

a1=a2+*r=2*a2

now equating force on plank

5.40-2*fp=6.6*a1

let the friction force due to ground and plank be in forward direction

fp+fg=2*a2

torque balancing eq

(fp-fg)*r=(2*r2/2)*

fp-fg=a2

solving the above equations

fp=3*a2/2

5.40=6.6*a1+3*a2

5.40=13.2*a2+3*a2

a2=.33 m/s2

a1=2*.33=.66m/s2

fp=1.5*.33=.495N

fg=.5*.33=.165N

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