Question

T 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? Serway, Raymond A.. College Physics (p. 114). Brooks Cole. Kindle Edition.

Answer 1

a= 1.59 m/s² : Magnitude of the acceleration

β = 65.22° (north of east) : Direction of the acceleration

Step-by-step explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

The acceleration is presented in the direction of the resultant force applied .

Calculation of the resultant forces (R)

`R=\sqrt{(F_(N))^(2) +(F_(E))^(2) }`

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

R= 429.5 N

We apply the formula (1) to calculate the magnitude of the acceleration(a) :

∑F = m*a , m= 270 kg

R= m*a

429.5 =270*a

`a=(429.5)/(270)  (m)/(s^(2) )`

a= 1.59 m/s²

Calculation of the direction of the acceleration (β)

`\beta = tan^(-1) ((F_(N) )/(F_(E)))`

`\beta = tan^(-1) ((390 )/(180))`

β = 65.22° (north of east)