WAEC 2010 · Paper 2 · Q18

(a)
Two forces $(3\mathbf i + 5\mathbf j)$ N and $(-2\mathbf i + 3\mathbf j)$ N act on a body of mass 2 kg and cause it to move. Find, correct to two decimal places, the magnitude of the: (i) resultant force acting on the body; (ii) acceleration of the body; (iii) change in velocity, if the forces acted on the body for 5 seconds.
Separate values with commas, e.g. 3, −2
(b)
A uniform beam of length 100 cm and mass 1.5 kg is placed on a pivot which is 20 cm from one end. A vertical force, $T$ , is applied upwards 5 cm from the other end to keep the beam in equilibrium. Calculate the: (i) magnitude of $T$ ; (ii) reaction at the pivot. [Take $g = 10$ m s $^{-2}$ ]
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Add the forces: $(3 - 2)\mathbf i + (5 + 3)\mathbf j = \mathbf i + 8\mathbf j$ .
Magnitude: $\sqrt{1^2 + 8^2} = \sqrt{65} \approx 8.06$ N.

Change in velocity $= at$
$= \dfrac{\sqrt{65}}{2} \times 5$
$\approx 20.16$ m s $^{-1}$ .

The weight is $1.5 \times 10 = 15$ N at the middle, 50 cm from the end: that is $50 - 20 = 30$ cm from the pivot.
$T$ acts 95 cm from the first end: $95 - 20 = 75$ cm from the pivot, on the same side as the weight.