Page 39 - Schaum's Outline of Theory and Problems of Applied Physics

24 VECTORS [CHAP. 2

F x = F cos q
F = F sin q
y
F y
F
q
F x

Fig. 2-13 Fig. 2-14
SOLVED PROBLEM 2.10

The man in Fig. 2-13 exerts a force of 100 N on the wagon at an angle of θ = 30 above the horizontal.
◦
Find the horizontal and vertical components of this force.
The magnitudes of F x and F y are, respectively,
◦
F x = F cos θ = (100 N)(cos 30 ) = 86.6N
F y = F sin θ = (100 N)(sin 30 ) = 50.0N
◦
We note that F x + F y = 136.6 N although F itself has the magnitude F = 100 N. What is wrong? The answer is
that nothing is wrong. Because F x and F y are just the magnitudes of the vectors F x and F y , it is meaningless to add
them. However, we can certainly add the vectors F x and F y to find the magnitude of their resultant F. Because F x
and F y are perpendicular,

2 2 2 2
F = F + F = (86.6N) + (50.0N) = 100 N
x
y
as we expect.
SOLVED PROBLEM 2.11
A woman in a car on a level road sees an airplane traveling in the same direction that is climbing at an
angle of 30 above the horizontal. By driving at 110 km/h she is able to stay directly below the airplane.
◦
Find the airplane’s velocity.
The car’s velocity is equal to the horizontal component v x of the airplane’s velocity v, as in Fig. 2-15. Since
v x = v cos θ

the airplane’s velocity v is
110 km/h
v x
v = = = 127 km/h
cos θ cos 30 ◦

Fig. 2-15