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const2013 [10]
4 years ago
9

While camping in Denali National Park in Alaska, a wise camper hangs his pack of food from a rope tied between two trees, to kee

p 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?

Physics
1 answer:
Crank4 years ago
4 0

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 <em>T</em> 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.




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