Question

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east. If the boat (including its crew) has a mass of 260 kg, what are the magnitude and direction of its acceleration? magnitude m/s2 direction ° north of east

Answer 1

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is  $\theta = 32.5 6^o$ north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  $F_{sail} = (330 ) \ N \ north$

   The force exerted by water is  $F_{keel} = (210 ) \ N \ east$

      The mass of the boat(+ crew) is  $m_b = 260 \ kg$

Now Force is mathematically represented as

      $F = ma$

Now the acceleration towards the north is mathematically represented as

      $a_n = \frac{F_{sail}}{m_b}$

substituting values

       $a_n = \frac{330 }{260}$

      $a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

       $a_e = \frac{F_{keel}}{m_b }$

substituting values

      $a_e = \frac{210}{260}$

      $a_e =0.808 \ m/s^2$

The resultant acceleration is  

      $a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

     $a_r = \sqrt{(0.808)^2 + (1.269)^2}$

      $a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        $tan \theta = \frac{a_e}{a_n}$

       $\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

     $\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

    $\theta = tan ^{-1} [0.636 ]$

   $\theta = 32.5 6^o$