Problem

(I) A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end (Fig. 4-45). Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 N, (h) 60.0 N. and (c) 90.0 N.

Solution 19P: We have find out the force exerted by the table on the box under the application of weight in various denomination, in a pulley-rope system. Step 1 of 5 Concept: Newton’s second law: The force F acts on mass m produces an acceleration a in the object. Mathematically, F=ma Newton’s third law: To every action, there is an equal and opposite reaction. Step 2 of 5 Figure below shows the free body diagram of box and and hanging box in various cases. Normal reaction on the box= F N Weight of box on the table = FG1 = 77 N Tension in the string = F T Weight of the hanging box = F G2 Taking positive direction along upward direction and negative direction along downward direction. Step 3 of 5 A] When the weight of the hanging box = F = 30 N G2 As the weight of the hanging box is less than the weight of the box placed on the table as shown in fig(2), both the boxes will not move at all, therefore according to Newton’s second law, the acceleration of the system would be a = 0, as the sum of forces = 0.However, tension would be developed in the rope. From fig(2) of the hanging box, F T = F G2 = 30 N From fig(1) of the box on the table,F + F = F T N G1 FN = F G1 F T FN = 77 30 F = 47 N N Thus, normal force exerted by the table on the boxNis= 47 N