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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

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?

To determine
The magnitude and direction of acceleration of the boat.

Explanation

Section 1:

To determine: The magnitude and direction of the resultant force.

Answer: The resultant force has a magnitude of 429.53 N directed at an angle of 65.22ο north of east.

Explanation:

Given info: The force exerted by the wind on the sails is 390 N north. The force exerted by water is 180 N east.  The mass of the boat is 270 kg.

The forces act perpendicular to each other. Therefore, the formula to calculate the magnitude of resultant force is,

Fr=FE2+FN2

  • FN is the force along the north direction.
  • FE is the force along the east direction.

Substitute 390 N for FN and 180 N for FE in the above expression to get Fr .

Fr=(180N)2+(390N)2=429.53N

The magnitude of the resultant force is 429.53 N.

The direction of the resultant force is,

θ=tan1(FNFE)

Substitute 390 N for FN and 180 N for FE in the above expression to get θ

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