(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 5 kg moves with initial velocity (frac{1}{2} m/s) and final velocity (frac{3}{4} m/s ). Find the magnitude of its change in momentum.
Explanation
(i) (T_{1} = begin{pmatrix} -T_{1} cos 40° \ T_{1} sin 40° end{pmatrix} = begin{pmatrix} -0.7660T_{1} \ 0.6428T_{2} end{pmatrix})
(T_{2} = begin{pmatrix} T_{2} cos 30° \ T_{2} cos 30° end{pmatrix} = begin{pmatrix} 0.8660T_{2} \ 0.5000T_{2} end{pmatrix})
(W = begin{pmatrix} 0 \ -65 end{pmatrix})
(ii) P is at rest, the vector equation is
(T_{1} + T_{2} + W = 0)
(iii) (begin{pmatrix} -0.7660T_{1} \ 0.6428T_{1} end{pmatrix} + begin{pmatrix} 0.8660T_{2} \ 0.5000T_{2} end{pmatrix} + begin{pmatrix} 0 \ -65 end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix})
(0.7660T_{1} = 0.8660T_{2} … (1))
(0.6428T_{1} + 0.5000T_{2} – 65 = 0 … (2))
From (1), (T_{1} = frac{0.8660T_{2}}{0.7660} … (3))
Put (3) in (2) :
(0.6428(frac{0.8660T_{2}}{0.7660}) + 0.5000T_{2} = 65 )
(0.7627T_{2} + 0.5000T_{2} = 65)
(1.2267T_{2} = 65 implies T_{2} = frac{65}{1.2267})
(T_{2} = 52.988N approxeq 53.0N) (to 1 d.p)
From (3), (T_{1} = frac{0.866 times 52.988}{0.766} = 59.905N)
(approxeq 59.9N) (to 1 d.p)