physics questions on vectors?

Question

1.) A horizontal and a vertical force combine to give a resultant force of 10 N acts in a direction 40 degress above the horizontal. Find the magnitude of the horizontal and vertical force.

ans: Fh=7.7 N Fv= 6.4 N

2.) A sailboat cannot sail directly toward the wind but must "tack" back and forth at a certain angle with respect to the direction from which the wind is blowing. Which sailboat has the greater component of velocity to windward, the Alpha whose velocity is 5.0 km/h at and angle of 40 degrees away from the wind, or the Beta, whose velocity is 6.0 km/h at an angle of 50 degrees away from the wind?

ans:alpha= 3.83 km/h beta= 3.86 km/h

like how do u get these problems b/c i worked all night tryin to figure these out.

Answer 1

For question 1 the rule is to use the cosine of the angle between. So since the force is 40 deg from the horizontal, you take the resultant and multiply it by the angle from the horizontal to get the horizontal component. Or:

Fh = 10 * cos(40 deg) = 7.7

and for the vertical component:

Fv = 10 * sin(40 deg) = 6.4


It's pretty much the same thing for question 2. You have the velocity and the angle from the direction of the wind, so for Alpha you do:

alpha = 5 * cos(40 deg) = 3.83

and for Beta:

beta = 6 * cos(50 deg) = 3.86

Answer 2

1. Fh = F cos(angle); Fv = F sin(angle); where F = 10 N and angle = 40 degrees, and Fh and Fv are the horizontal and vertical forces. You do the math.

2. Vh(alpha) = 5 kmph = V cos(40); Vh(beta) = 6 = V cos(50); so that Vv(alpha) = ? = V sin(40) and Vv(beta) = ? = V sin(50) Therefore:

Vv(alpha) = (5/cos(40)) X sin(40) = 5 tan(40) and
Vv(beta) = 6 tan(50). Again you can do the math. But you can see that the beta boat has the greater vertical velocity Vv, which is the boat's windward velocity.

Answer 3

1. You must use trigonometry to break the forces into horizontal and vertical:

10 cos 40 = 7.66

10 sin 40 = 6.43

2. Same thing:

5 cos 40 = 3.83

6 cos 50 = 3.86

BE SURE YOUR CALCULATOR IS SET TO DEGREES, NOT RADIANS!!!

Here is how I remember it:

Some old horse caught another horse taking oats away. Translated:

Sin Theta = opposite/hypoteneuse (some old horse)

Cos Theta = adjacent/hypoteneuse (caught another horse)

Tan Theta = opposite/adjacent (taking oats away)

Now. Let's go back to our original problems:

1. Cos 40 = adjacent / 10

So balancing the equation:

10 Cos 40 = adjacent

For the next one:

Sin 40 = opposite / 10

So balancing the equation:

10 Sin 40 = opposite

Problem #2 is solved in the exact same manner. The key to the proper use of trigonometry is deciding how to draw the triangle. Usually there is only one correct way and many wrong ways.