A plank with a mass $M = 6.00 k g$ rests on top of two identical solid cylindrical rollers that have $R = 5.00 c m$ and $m = 2.00 k g$ (figure). The plank is pulled by a constant horizontal force $\to F$ of magnitude $6.00$ $N$ 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. All surfaces are rough.

The initial acceleration of the plank at the moment when the rollers are equidistant from the ends of the plank is
$m / s^{2}$ .

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Solution

Let $f_{t}$ be the force of friction between plank and the cylinder and $f_{b}$ be the force of friction between cylinder and the ground.

For the plank, $\sum F_{x} = m a_{x}$
$6.0 N - 2 f_{t} = (6.00 k g) a_{p}$ . .........(i)
Now, as there is no slipping between the plank and the cylinders we can say
$a_{p} = a_{c} + R α$ ...........(ii)
Also there is no slipping between ground and the cylinders so,
$a_{c} = R α$ ............(iii)
By using (ii) and (iii) we can say
each roller has acceleration $\frac{a_{p}}{2}$ and angular acceleration
$α = \frac{a_{p} / 2}{5.00 c m} = \frac{a_{p}}{0.100 m}$ Then for each cylinders; $\sum F_{x} = {m a}_{x}$
$f_{t} - f_{b} = (2.00 k g) \frac{a_{p}}{2} . . . . (i v)$
Now applying torque equation $\sum τ = I α$
$f_{t} (5 c m) + f_{b} (5 c m) = \frac{1}{2} (2 k g) (5 c m)^{2} \frac{a_{p}}{10 c m}$ So $f_{t} + f_{b} = (\frac{1}{2} k g) a_{p}$ ..........(v)
Substituting the value for 2 $f_{t}$ into equation (i) it gives
$6.00 N - (1.50 k g) a_{p} = (6.0 k g) a_{p}$
$a_{p} = \frac{6.00 N}{7.50 k g} = 0.80 m / s^{2}$

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