Sol. As the string is unstrechable, when the center of C move up the inclined plane a distance Δ x , the 2kg mass will also drop Δ x. However, as an additional length Δ x of the string is also released in the process, the 2 kg mass will actually drop 2 Δ x . Thus when the 2kg mass moves down one meter , C will move up the inclined plane 0.5 m, or vertically up 0.5 sin 30º = 0.25 m. Ans.

The force involved are shown in above fig. The above effect means that for the 2kg mass we have

2m = mg – F.

For the cylinder we have

M = F + f – Mg sin 30º

I = (F – f) R,

Where I = MR 2 . Furthermore, as the cylinder

rolls without slipping we also have

= R

The above equations give

= g = –0.0435g

= –0.426 ms –2 Ans.

Thus the acceleration has magnitude 0.426 ms –2 and acts downward along the inclined plane

f = 4M ( + g) = 40 × 0.57g = 23.0N. Its direction is upward along the inclined plane.