yvette hangs a 2.4kg bird feeder in the middle of a rope tied between two trees, the feeder creates a
tension of 480n in eachside of rope. a) if the two trees are 4m apart, how much will the rope sag in the
center?b) if a bird lands on the feeder, will this have any effect on the tension in the rope? explain.

In the y direction, there are three forces acting on the feeder. Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.

Apply Newton's second law to the feeder in the y direction.

∑F = ma

2Ty − mg = 0

Ty = mg/2

Let's say the distance the rope sags is d. The trees are 4m apart, so the feeder is 2m horizontally from either tree. Using Pythagorean theorem, we can find the length of the rope on either side:

L² = 2² + d²

L = √(4 + d²)

Using similar triangles, we can write a proportion using the forces and distances.

Ty / T = d / L

Substitute:

(mg/2) / T = d / √(4 + d²)

Solve for d:

Td = mg/2 √(4 + d²)

T² d² = (mg/2)² (4 + d²)

T² d² = (mg)² + (mg/2)² d²

(T² − (mg/2)²) d² = (mg)²

d² = (mg)² / (T² − (mg/2)²)

d = mg / √(T² − (mg/2)²)

Given m = 2.4 kg and T = 480 N:

d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)

d = 0.049 m

b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.