Figure P 5.19 shows the horizontal forces acting on a...

Figure $\mathrm{P} 5.19$ shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from at point straight above its mast. At the particular speed of the sailboat, the water exerts a 220 -N drag force on its hull and $\theta=40.0^{\circ} .$ 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 $\theta=$ $40.0^{\circ}$ north of east and the $y$ direction as $\theta=40.0^{\circ}$ west of north. (c) Compare your solutions to parts (a) and (b). Do the results agree? Is one method significantly easier?

Figure $\mathrm{P} 5.19$ shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from at point straight above its mast. At the particular speed of the sailboat, the water exerts a 220 -N drag force on its hull and $\theta=40.0^{\circ} .$ 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 $\theta=$ $40.0^{\circ}$ north of east and the $y$ direction as $\theta=40.0^{\circ}$ west of
north. (c) Compare your solutions to parts (a) and (b). Do the results agree? Is one method significantly easier?

Key Concepts

Newton's Second Law

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In problems where the velocity is constant, the acceleration is zero, implying that the vector sum of all forces must vanish. This fundamental law is used to set up equations in each coordinate direction, ensuring that the forces balance to yield equilibrium.

Force Equilibrium

Force equilibrium means that an object experiences no net acceleration when the sum of forces in every direction equals zero. In this case, since the sailboat is moving at constant velocity, the forces acting on it must cancel each other perfectly. This concept is crucial for writing the component equations along each chosen coordinate axis.

Vector Component Decomposition

Breaking forces into components along chosen axes allows us to analyze their effects independently in each direction. By resolving forces such as the wind thrust or water pressure into horizontal (x) and vertical (y) components, we can apply Newton's second law along each axis separately, simplifying the process of finding unknown magnitudes.

Choice of Coordinate Systems

Selecting an appropriate coordinate system can simplify the resolution of forces by aligning the axes with the directions of some forces, thus reducing the number of nonzero components. In the sailboat problem, using two different coordinate systems provides an opportunity to compare the ease of analysis and reinforces the idea that physical predictions are independent of the choice of coordinates.

Transcript

00:01 In this question we have a sailboat that is experiencing a few forces.

00:06 We have a normal force that is acting towards the east, sorry west.

00:15 So the direction is up is north.

00:20 And then we have a pulling force from the wind acting on a mast on the sailboat.

00:31 And there's acting 40 degrees north of the east direction.

00:40 So we want to use two different coordinates to find the answer, and see which one will be easier.

00:49 So the first coordinate we're going to use is just based on the actual compass, with the north being our y -axis, and the east being our x -axis.

01:00 So in this case, because the sailboat is moving with constant velocity, it means that all the net forces in both the x and the y direction must add up to be 0.

01:21 For the x direction, we have the x component from the p as well as the x component from the normal force.

01:31 This is just an must be equals to p times cosine 40 degrees their magnitude must be the same because they are in opposite directions so that they will cancel each other out now we consider the y direction so the y direction only comes from the drag force 220 noughts as well as the y component of p so the drag force must be equal in magnitude to the y component of p which is just p sine 40 degrees.

02:21 Now with these two equations we can solve pretty easily, right? one way is to divide equation 1 by equation 2 and you will get n over 220 nuitons goes 2, cosine 4 divided by, sine 40.

02:50 And you should get n to be 262 newtons.

02:58 And you can substitute back into equation 1.

03:00 Now you should be able to find p, should be 3 ,4 2 mutants.

03:07 Now the other coordinate system that we are asked to use is when the xx axis is angle 40 degrees in this direction...

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