The figure below shows the horizontal forces acting on a sailboat moving north a

Question

The figure below shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from a point straight above its mast. At the particular speed of the sailboat, the water exerts a F with arrow = 400-N drag force on its hull and = 35.0°. For each of the situations (a) and (b) described below, write two component equations representing Newton's second law. Then solve the equations for P (the force exerted by the wind on the sail) and for n (the force exerted by the water on the keel).

(a) Choose the x direction as east and the y direction as north.

(b) Now choose the x direction as = 35.0° north of east and the y direction as = 35.0° west of north.

(c) Compare your solutions to part (a) and (b). Do the results agree? Yes No

Explanation / Answer

from netwons second law

Fnet = ma

speed is constant , so acceleration a = 0

along

y direction

Fnety = 0

P*sintheta - F = 0

P*sintheta = F

P = F/sintheta

P = 400/sin35

P = 697.4 N

along x axis

Fnetx = 0

P*cos35 - n = 0

n = P*cos35 = (F/sin35)*cos35

n = P/tan35

n = 400/tan35

n = 571.25 N

=========================================

(b)

along north

Fnet = 0

P*cos(90-35) - F = 0

P = F/cos(90-35)

P = 400/sin35

along east

Fnet = 0

P*sin(90-35) -n = 0

n = P*cos35 = (F/sin35)*cos35)

n = F/tan35 = 571.25 N

c)

the results agree

Yes

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