A sailboat is moving at a constant velocity due north, as...

A sailboat is moving at a constant velocity due north, as depicted in the figure. The water exerts a drag force on the boat, where the magnitude of F is 300 N and ? = 35.0°. (a) If +x is east and +y is north, find the magnitudes of n and p. (Give your answers in N.) n = p = (b) If +x is ? = 35.0° north of east and +y is ? = 35.0° west of north, find the magnitudes of n and p. (Give your answers in N.) n = p = (c) Do your answers to parts (a) and (b) agree? Yes No

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00:01 First consider sub -8.

00:02 It is given that the sailboat is moving with a constant speed in the north directions, which implies the sailboat is moving with the constant speed in the y -axis.

00:12 Since the speed is constant, the net force acting on the sailboat in both x and y directions will be zero.

00:20 Therefore, sigma fy is equal to zero.

00:23 Here, sigma fy is the net force acting on the sailboat in the y direction.

00:28 The equation for sigma fy is given by sigma fy is equal to p sine theta minus f.

00:38 Equating this equation to zero gives, p sine theta minus f is equal to zero.

00:46 Or p sine theta is equal to f or p is equal to f or p is equal to f by sine theta.

00:55 Substituting the values in the equation gives, p is equal to 300 newton divided by 535 .0 degree, which is equal to 523 .03 newton.

01:16 Now, the next force acting on the sailboat in x direction also must be equal to zero.

01:23 This gives sigma fx is equal to zero.

01:26 And the individual force acting on the sailboat in the x direction are the cause component of the the force p in the plus x direction and the normal force in the minus x direction.

01:38 Therefore, sigma fx is equal to p cost theta minus n.

01:43 So equating this to zero gives, p cost theta minus n is equal to zero...

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