Q#694: [Dynamics > Equilibrium] (Past Exam Paper – June 2011 Paper 23 Q2)
Climber is supported by a rope on a vertical wall, as shown in Fig.1.
Weight W of the climber is 520 N. The rope, of negligible weight, is attached to climber and to a fixed point P where it makes an angle of 18° to the vertical. Reaction force R acts at right-angles to the wall.
The climber is in equilibrium.
(a) State conditions necessary for the climber to be in equilibrium.
(b) Complete Fig.2 by drawing a labelled vector triangle to represent the forces acting on the climber.
(c) Resolve forces or use vector triangle to calculate
(i) tension T in the rope,
(ii) reaction force R.
(d) Climber moves up the wall and the angle the rope makes with the vertical increases. Explain why magnitude of the tension in the rope increases.
Solution 694:
(a)
Resultant moment = zero / sum of clockwise moments = sum of anticlockwise moments
Resultant force = 0
(b)
shape and orientation correct and forces labelled and arrows correct
angles correct / labelled
{Tension T in rope = Weight W + Reaction R}
(c)
(i) {The vertical forces should be balanced.}
T cos18° = W
Tension T = 520 / cos18° = 547 N (Scale diagram: ± 20 N)
(ii) {The horizontal forces should be balanced.}
Reaction force R = T sin18° = 169 N (Scale diagram: ± 20 N)
(d) The angle θ is larger, hence cos θ is smaller. Tension T = W / cos θ. Hence the tension T is larger.
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