The two conditions of equilibrium: boards, rods and strings
Week 8 · Equilibrium of coplanar forces: the two conditions (zero resultant and zero moment), a board on two supports, and a rod on two strings · 3 lessons
3
A rod hung from two vertical strings: where must a weight hang so that one tension is double the other?
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 2, Part 1
GoalFind the tensions in two vertical strings that hold a rod, and decide where to hang a weight so that one tension is a given multiple of the other.
New words
tension · the pull of a string on the rod, along the stringvertical strings · the two tensions are parallel forces pointing up, just like the reactions of supports
Explanation
A rod hung by two vertical strings is the same problem as a rod on two supports, with the tensions TA and TB in place of the reactions: they pull up, the weights pull down. Use the same two equations: TA + TB = total weight and one moment equation about an end. If the question says ‘one tension is double the other’, write TA = 2 TB, use the first equation to get both tensions from the total weight, and then the moment equation to find the unknown distance. The weight must hang nearer to the string that carries more. The tensions of the two strings are always positive: if an equation gives a negative tension, the string would have to push, which is impossible, so the rod would tip.
Worked examples
A uniform rod AB, 100 cm long and weighing 50 N, hangs level from two vertical strings at A and B. A weight of 100 N hangs at x cm from A. Find x if TA = 2 TB. Up and anticlockwise are positive. Total weight 50 + 100 = 150 N so TB = 50 N and TA = 100 N. Moments about A: 100 TB − 50 × 50 − 100x = 0, that is 5000 − 2500 = 100x and x = 25 cm. The weight hangs nearer A, the string with the bigger tension.
A uniform rod AB, 120 cm long and weighing 60 N, hangs level from two vertical strings at A and B. A load of 90 N hangs y cm from B. Find y when TB = 3 TA. Up and anticlockwise are positive. Total weight 150 N, so TA = 37.5 N and TB = 112.5 N. Moments about A: 120 × 112.5 − 60 × 60 − 90 × (120 − y) = 0, so 13500 − 3600 − 10800 + 90y = 0 and y = 10 cm from B.
Class activity
Hang a light ruler level from two loops of thread, one at each end, held by two friends at the same height. Clip a pencil sharpener (a light load) on the ruler near one end, then near the middle. The friend nearer the load feels the thread pull more. Use light loads only (under 0.2 kg) and keep your hands clear of the ruler.
Practice
1
A uniform rod AB, 60 cm long and weighing 40 N, hangs level from vertical strings at A and B. A weight of 80 N hangs 15 cm from A. Up and anticlockwise are positive. Find TA and TB.
About B: 60 TA = 40 × 30 + 80 × 45 = 4800, so TA = 80 N and TB = 120 − 80 = 40 N.
2
For the same rod and the same 80 N weight, how far from A must the weight hang so that TA = 3 TB? Up and anticlockwise are positive.
Total 120 N, so TB = 30 N. About A: 60 × 30 = 40 × 30 + 80x, so x = 7.5 cm.
3
A uniform cleaner’s plank of weight 300 N and length 6 m hangs level from vertical ropes at its two ends. A man of weight 600 N stands 1.5 m from one end. Up and anticlockwise are positive. Find the tension in each rope.
Let the man be 1.5 m from end A. About B: 6 TA = 300 × 3 + 600 × 4.5 = 3600, so TA = 600 N (the rope nearer the man) and TB = 900 − 600 = 300 N.
4
A uniform rod AB, 80 cm long and weighing 30 N, hangs level from vertical strings at A and at C, where AC = 60 cm. There is no other load. Up and anticlockwise are positive. Find TA and TC.
The weight acts 40 cm from A, 20 cm to the left of C. About C: −60 TA + 20 × 30 = 0, so TA = 10 N and TC = 30 − 10 = 20 N.
5
A uniform rod of weight 50 N hangs level from two vertical strings at its ends. A weight w is hung at the midpoint. Show that the two tensions stay equal, whatever w is.
Both weights act at the midpoint, so by symmetry the moments about the midpoint give TA = TB, and TA + TB = 50 + w gives TA = TB = (50 + w) / 2.
