CP? A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?
Solution 3E To solve this question, let us have a look at the figure below where various forces and their components are shown for the system. Let the tension in the rope is T and acts as shows above. From the figure, sin x = 3.5 x = sin 1(2.0) 3.5 0 x = 34.7 Also, the horizontal force F H = Tsin x…..(1) Weight of the bag mg = Tcos x…..(2) (a) Dividing equation (1) by (2), F H mg = tan x F = mgtan x H F = 90.0 kg × 9.80 m/s × tan 34.7 0 H F H = 611 N Therefore, the approximate value of the horizontal force is 611 N. 0 (b) The height moved by the bag is = 3.5 3.5cos x = 3.5 3.5cos 34.7 = 0.62 m 2 Work done by the worker is = 0.62 m × 90 kg × 9.8 0m/s = 547 N The approximate work done by the worker is 547 N. The rope itself cannot change its position unless a force is applied on it. Therefore, work done by the rope is zero.