Theory question 116
Statics(a) An object P of mass 6.5kg is suspended by two light inextensible strings AP and BP. The strings make angles 50° and 60° respectively with the downward vertical. (i) Express the forces acting on P in component form. (ii) If P is at rest, write down the vector equation connecting all the forces. (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5kg moves with initial velocity (−1 / 2)ms−1 and final velocity (3 / 4)ms−1. Find the magnitude of its change in momentum. (Column vectors are written top / bottom.)
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(a) Take g = 10 ms−2: weight W = 65 N. Let T1 (in AP) and T2 (in BP)
(i) T1 = (−T1 sin 50° / T1 cos 50°), T2 = (T2 sin 60° / T2 cos 60°), W = (0 / −65)
(ii) T1 + T2 + W = 0
(iii) Horizontal: T1 sin 50° = T2 sin 60°; vertical: T1 cos 50° + T2 cos 60° = 65
Solving (or Lami's theorem, angle between strings 110°): T1 = 65 sin 60°/sin 110° ≈ 59.9 N, T2 = 65 sin 50°/sin 110° ≈ 53.0 N
(with g = 9.8: 58.7 N and 51.9 N)
(b) Change in momentum = m(v − u) = 5[(3 / 4) − (−1 / 2)] = 5(4 / 2) = (20 / 10) Ns
Magnitude = √(400 + 100) = √500 = 10√5 ≈ 22.4 Ns
Resolve horizontally and vertically; the horizontal parts of the two tensions must balance.