Theory question 117
Dynamics(a) Forces F1 = (−5, 4)N, F2 = (2, 5)N, F3 = (2, −1)N, F4 = (3, −5)N, act on a body, Find the: (i) resultant of these forces; (ii) fifth force that will keep the system in equilibrium.
(b) A body, moving at 20ms−1 accelerates uniformly at 2.5ms−2 for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest t seconds with a uniform retardation. If the ratio of acceleration to retardation is 3:4 (i) sketch the velocity - time graph of the journey; (ii) find t, (iii) find the total distance of the journey.
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(a)(i) R = (−5 + 2 + 2 + 3, 4 + 5 − 1 − 5) = (2, 3)N (magnitude √13 N)
(ii) Fifth force = −R = (−2, −3)N
(b)(i) Graph: line from 20 to 30 ms−1 over 0–4 s, horizontal at 30 ms−1 from 4 to 12 s, then down to 0
Speed after 4 s: 20 + 2.5 × 4 = 30 ms−1
(ii) Retardation = 2.5 × 4/3 = 10/3 ms−2; t = 30 ÷ 10/3 = 9 s
(iii) Distance = area: ½(20 + 30)(4) + 30 × 8 + ½(30)(9) = 100 + 240 + 135 = 475 m
The equilibrant is equal and opposite to the resultant; distance is the area under the velocity–time graph.