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Nov 2009 p42 q1
4057
A small block of weight 12 N is at rest on a smooth plane inclined at 40° to the horizontal. The block is held in equilibrium by a force of magnitude P N. Find the value of P when
the force is parallel to the plane as in Fig. 1,
the force is horizontal as in Fig. 2.
Solution
(i) When the force is parallel to the plane, resolve the forces parallel to the plane. The component of the weight acting down the plane is given by:
\(W \sin 40^{\circ} = 12 \sin 40^{\circ}\)
Equating this to the force P, we have:
\(P = 12 \sin 40^{\circ}\)
Calculating gives:
\(P = 7.71 \text{ N}\)
(ii) When the force is horizontal, resolve the forces horizontally and vertically. The horizontal component of P is \(P \cos 40^{\circ}\) and the vertical component of the weight is \(W \sin 40^{\circ}\). Equating the horizontal component of P to the vertical component of the weight gives:
