(a) The position vectors of points A, B and C are (i + 5j , 3i + 9j) and (-i + j) respectively. (i) Show that points A, B and C are collinear; (ii) Determine the ratio (|AB| : |BC|).
(b) A uniform beam XY of mass 10 kg and length 24m is hunged horizontally from a cross bar by teo vertical inextensible strings, one attached to X and the other at a point M, 4m away from Y. A mass of 50kg is suspended at a point N which is 8m from X. If the system remains in equilibrium, calculate the tensions in the strings.
Explanation
(a)(i) Position vector of A, (overrightarrow{OA} = i + 5j)
Position vector of B, (overrightarrow{OB} = 3i + 9j)
Position vector of C, (overrightarrow{OC} = -i + j)
The points A, B and C are collinear if (|overrightarrow{AB}| = k|overrightarrow{BC}|), where k is a constant.
If k > 0, then (overrightarrow{AB}) and (overrightarrow{BC}) are in the same direction.
If k < 0, then they are in opposite directions, in this case they are A, C, B.
(overrightarrow{AB} = overrightarrow{AO} + overrightarrow{OB} = – overrightarrow{OA} + overrightarrow{OB})
= (- i – 5j + (3i + 9j) = 2i + 4j = 2(i + 2j))
(overrightarrow{BC} = overrightarrow{BO} + overrightarrow{OC} = – overrightarrow{OB} + overrightarrow{OC})
= (- 3i – 9j + (-i + j) = – 4i – 8j = -4(i + 2j))
(overrightarrow{AB} = -frac{1}{2} overrightarrow{BC})
(therefore) A, B and C are collinear with (overrightarrow{AB}) and (overrightarrow{BC}) in opposite directions.
(ii) (|overrightarrow{AB}| : |overrightarrow{BC}| = 1 : 2)
(b)
We take moments about X.
Clockwise moment = (500 times 8 + 100 times 12 = 5200N)
Anticlockwise moment = (T_{2} times 20)
From the principles of moment,
(20T_{2} = 5200N implies T_{2} = 260N)
For vertical equilibrium,
(T_{1} + T_{2} = 600N)
(therefore T_{1} = 600 – T_{2} = 600 – 260 =340N)
The tensions in the strings are 340N and 260N respectively.