Theory question 116
Statics(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings XP and YP. The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensions in the strings. [Take g = 10 ms−2]
(b) The height h metres, of a ball thrown into the air is 2 + 20t + kt2, after t seconds. If it takes 2 seconds for the ball to reach its highest point, find: (i) the value of k; (ii) its highest point from the point of throw.
Show the model answer
(a) Weight = 150 N. Angle between the strings = 100°; angle between XP and the weight = 120°, between YP and the weight = 140°
Lami: TXP/sin 140° = TYP/sin 120° = 150/sin 100°
TXP = 150 × 0.6428/0.9848 ≈ 97.91 N
TYP = 150 × 0.8660/0.9848 ≈ 131.91 N
(b)(i) dh/dt = 20 + 2kt = 0 at t = 2, so k = −5
(ii) h(2) = 2 + 40 − 20 = 22 m above the reference level, i.e. 20 m above the point of throw (where h = 2)
Use Lami's theorem (or resolve horizontally and vertically); the highest point is where dh/dt = 0.