A plank with a mass M = 6.10 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.10 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force F 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.
(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.)
F = ma
I = (1/2)mR2 (Moment of inertia for a cylinder)
The Attempt at a Solution
From f = ma, I understand that the net force pulling the system is equal to the net inertial mass multiplied by the acceleration of the system.
Net force would be equal to F pulling minus force of friction giving me the equation:
F - Ff = ma
In order to find Ff I used the T=I∂ equation
(fp - fg)R = (1/2 * mR2) * (a/R)
Simplifying this results in
(fp - fg) = (1/2)ma
which I can plug back into F = ma netting me
F - (1/2)ma = mneta
So I plug in all my values:
4.8 - a = (6.1 + (2)(1/2)(2)(.041^2))a
and I get a = .676 for part (a)
and a = .338 for part (b)
Both of these answers are close, but wrong. The correct answers for a and b are .632 and .316, respectively. What am I doing wrong/missing?
Then, I'm unsure where to begin for part (c). If I plug in acceleration back into my net torque equations, I'm left with 2 unknowns and 1 equation.