A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

Summing vertical forces to zero

Tcosθ - mg = 0

T = 90.0(9.81) / cos34.85 = 1,075.85 N

Summing horizontal forces to zero

F - Tsinθ = 0

F = 1075.85sin34.85 = 1075.85(2.0/3.5) = 614.772... ≈ 615 N

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