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

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

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

Step-by-step 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 = (F_(sail))/(m_b)`

substituting values

`a_n = (330 )/(260)`

`a_n = 1.269 \ m/s^2`

Now the acceleration towards the east is mathematically represented as

`a_e = (F_(keel))/(m_b )`

substituting values

`a_e = (210)/(260)`

`a_e =0.808 \ m/s^2`

The resultant acceleration is

`a_r = √(a_e^2 + a_n^2)`

substituting values

`a_r = √((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 = (a_e)/(a_n)`

`\theta = tan ^(-1) [(a_e)/(a_n ) ]`

substituting values

`\theta = tan ^(-1) [(0.808)/(1.269 ) ]`

`\theta = tan ^(-1) [0.636 ]`

`\theta = 32.5 6^o`