(a) Forces (F_{1} = begin{pmatrix} -5 & 4 end{pmatrix} N; F_{2} = begin{pmatrix} 2 \ 5 end{pmatrix} N; F_{3} = begin{pmatrix} 2 & -1 end{pmatrix} N) and (F_{4} = begin{pmatrix} 3 & -5 end{pmatrix} N) act on a body. Find the :
(i) resultant of these forces ; (ii) fifth force that will keep the body in equilibrium.
(b) A body moving at 20 ms(^{-1}) accelerates uniformly at 2.5 ms(^{-2}) for 4 seconds. It continues the journey at the speed for 8 seconds, before coming to rest in t seconds with a uniform retardation. If the ratio of the acceleration to the retardation is 3 : 4,
(i) sketch the velocity- time graph of the journey ; (ii) find t ; (iii) find the total distance of the journey.
Explanation
(a) (i) (F_{1} = begin{pmatrix} -5 & 4 end{pmatrix} N; F_{2} = begin{pmatrix} 2 \ 5 end{pmatrix} N; F_{3} = begin{pmatrix} 2 & -1 end{pmatrix} N ;F_{4} = begin{pmatrix} 3 & -5 end{pmatrix} N)
(R = begin{pmatrix} -5 & 4 end{pmatrix} N + begin{pmatrix} 2 \ 5 end{pmatrix} N + begin{pmatrix} 2 & -1 end{pmatrix} N + begin{pmatrix} 3 & -5 end{pmatrix} N = begin{pmatrix} 2 \ 3 end{pmatrix} N)
(ii) Let the fifth force be F.
(F + begin{pmatrix} 2 \ 3 end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix})
(F = begin{pmatrix} -2 \ -3 end{pmatrix})
(b) (i)
Acceleration = a = 2.5 ms(^{-2}),
Let retardation be r ms(^{-2}).
(a : r = 3 : 4)
(therefore r = frac{4}{3} times 2.5 = 3frac{1}{3} ms^{-2})
(ii) We use the formula (v = u + at) for the retardation period.
(v = 0 m/s ; u = 30 m/s ; a = -3frac{1}{3} m/s^{2} ; t = ?)
(0 = 30 – frac{10}{3} t )
(t = frac{30 times 3}{10} = 9 secs)
(iii) Total distance = area under graph.
(A_{1} = frac{1}{2} (4)(20 + 30) = 100m)
(A_{2} = 30 times 8 = 240m)
(A_{3} = frac{1}{2} (9) (30) = 135m)
Total distance = 100m + 240m + 135m = 475m.