(A) - look, then, as Newton's third law states, the equal and opposite motions cancel each other out. Plus, they act at almost the same point, so their moments( Moment of Force = Force X Radius of action) will be the same as well. In both cases, Radius of action (r) = l/2 (l is length of pencil).

So, M1-M2 = F1Xr1 - F2Xr2 = F(r1-r2) = 0.

Net force, is clearly, = 0. The pencil will experience no motion at all.

(B) Now, since you did not mention the points of their action, I can only assume that they're acting at exactly the end points of the pencil.

They will thus, produce a **couple**. The moment of that couple = (F2-F1) x l. (l = length of pencil).

**NO FORCE** will act on the pencil, but a rotational motion will act, hence the pencil will move in a clockwise or anti-clockwise direction, depending on the nature of the forces mentioned.

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