Question

The force exerted by the wind on the sails of a sailboat is Fsail = 400 N north. The water exerts a force of Fkeel = 190 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

Establishing reference system with unit vectors.

• North = + j
• East = + i
• West = -i
• South = -j

Expressing as a vector each force acting on the boat.

• Force on the sail

$\vec{F_{Sail}}=400(j)Nw$

• Forces exerted by water:
• $\vec{F_{Keel}}=190(i)Nw$

The net Force is equal to:$\vec{F_{Net}}=(190i+400j)Nw$

Applying Newton's second Law on the sailboat

$\vec{F_{net}}=m*\vec{a}$

Where

• Net force $\vec{F_{Net}}=(190i+400j)Nw$
• Mass $m=260kg$
• Acceleration $\vec{a}$

Expression for acceleration:

$\vec{F_{net}}=m*\vec{a} \\ \vec{a}=\frac{\vec{F_{Net}}}{m}$

Evaluating numerically.

$\vec{a}=\frac{\vec{F_{Net}}=(190i+400j)Nw}{260kg} \\\vec{a}=(0.731i+1.54j)\frac{m}{s^{2}}$

The magnitude is given by: $|\vec{a}|=\sqrt{(0.731)^{2}+(1.54)^{2}} \\ |\vec{a}|=1.70\frac{m}{s^{2}}$

The magnitude of the acceleration is:$\boxed{|\vec{a}|=1.70\frac{m}{s^{2}}}$

The angle is given by

: $tan\theta=(\frac{ 1.54}{ 0.731})\\ \theta=tan^{-1}(\frac{ 1.54}{ 0.731}) \\ \theta=64.6^{0}$ Expressing respect to the East

The direction towards the east of the acceleration is: $\boxed{64.6^{0}\text{ north of east}}$