Question

. The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration?

Answer 1

1.59 m/s^2, 65.2°

Step-by-step explanation:

F1 = 390 N North

F2 = 180 N east

m = 270 kg

Net force is the vector sum of both the forces.

`F = \sqrt{F_(1)^(2)+F_(2)^(2)}`

`F = \sqrt{390^(2)+180^(2)}`

F = 429.53 N

Direction of force

tan∅ = F1 / F2 = 390 / 180 = 2.1667

∅ = 65.2°

The direction of acceleration is same as the direction of net force.

The magnitude of acceleration is

a = F / m = 429.53 / 270 = 1.59 m/s^2