A plank with a mass M = 6.80 kg rides on top of two identical, solid, cylindrica

Question

A plank with a mass M = 6.80 kg rides on top of two identical, solid, cylindrical rollers that have R = 5.10 cm and m 2.00 kg. The plank is pulled by a constant horizontal force F vector of magnitude 4.80 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. magnitude m/s^2 direction (b) Find the acceleration of the rollers at this moment. magnitude m/s^2 direction (c) What friction forces are acting at this moment? (Let f_p be the frictional force exerted by each roller on the plank, and let f_g be the rolling friction exerted by the ground on each roller.) F_p = N direction f_0 = N direction

Explanation / Answer

forces on the plank -2fg+F = Ma

2gt-2Mg = 0

F = 4.8 N

M = 6.8 kg

As each roller moves forward by the distance s, the plank moves twice as far. This means that the acceleration of each roller is half of the acceleration of the plank

vtop = 2Rw

s top = 2Rtheta

a top = 2R*alpha

The forces on the rollers is fg-fp=m*ar

(fg+fp)R = I*alpha = 0.5mR^2 *alpha = 0.5mR*ar

2 fg=1.5mar=3ar

fg=0.75*2*ar=1.5ar

fp=-0.25 mar

-3ar+4.8 = 6.8a = 13.6 ar

ar = 0.289 m/s^2

ap = 0.578 m/s^2

fp = -0.1445 N

fg = 0.4335 N

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