Question

While camping in Denali National Park in Alaska, a wise camper hangs his pack of food from a rope tied between two trees, to keep the food away from the bears. If the 50.-N bag of food hangs from the center of a rope that is 3.0 m long, and the rope sags 6.0 cm in the middle, what is the tension in the rope?

Answer 1

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,

$2T sin\theta = 50 N$     ......(1)

Calculate the value of sinθ  using the right angled triangles from the diagram.

$sin\theta =\frac{0.06 m}{1.5 m} =0.04$

Substitute the value of sinθ in equation (1) and simplify to obtain T.

$2T sin\theta = 50 N\\ T= \frac{50 N}{2 sin\theta}=\frac{50 N}{2*0.06} = 625 N$

Thus the tension in the rope is 625 N.