The tension in each of the ropes is 625 N.
Draw a free body diagram for the bag of food as shown in the attached diagram. Since the bag hangs from the midpoint of the rope, the rope makes equal angles θ with the horizontal. The tensions T in both the ropes are also equal.
Resolve the tension T in the ropes into horizontal and vertical components T cosθ and T sinθ respectively, as shown in the figure. At equilibrium,
[tex] 2T sin\theta = 50 N [/tex] ......(1)
Calculate the value of sinθ using the right angled triangles from the diagram.
[tex] sin\theta =\frac{0.06 m}{1.5 m} =0.04 [/tex]
Substitute the value of sinθ in equation (1) and simplify to obtain T.
[tex] 2T sin\theta = 50 N\\ T= \frac{50 N}{2 sin\theta}=\frac{50 N}{2*0.06} = 625 N [/tex]
Thus the tension in the rope is 625 N.
