Page 37 - Schaum's Outline of Theory and Problems of Applied Physics

22 VECTORS [CHAP. 2

SOLVED PROBLEM 2.8
In the triangle of Fig. 2-8, a = 30 mm and c = 90 mm. Find the values of b, θ, and φ.

2
2
2
2
b = c − a = (90 mm) − (30 mm) = 85 mm
a 30 mm
sin θ = = = 0.33
c 90 mm
θ = sin −1 0.33 = 19 ◦
◦
◦
◦
φ = 90 − θ = 90 − 19 = 71 ◦
VECTOR ADDITION: TRIGONOMETRIC METHOD
It is easy to apply trigonometry to find the resultant R of two vectors A and B that are perpendicular to each
other. The magnitude of the resultant is given by the Pythagorean theorem as

2
R = A + B 2
and the angle θ between R and A (Fig. 2-10) may be found from
B
tan θ =
A
by examining a table of tangents or by using a calculator to determine tan −1 B/A.

Fig. 2-10

SOLVED PROBLEM 2.9
Use trigonometry to solve Prob. 2.1.
From the vector diagram of Fig. 2-11 we see that A and B are the sides of a right triangle and R is its hypotenuse.
2
2
2
According to the Pythagorean theorem, the magnitudes A, B, and R are related by R = A +B . Hence the magnitude
R is equal to

2 2 2 2 2 2
R = A + B = (5km) + (10 km) = 25 km + 100 km

2
= 125 km = 11.2km
To find the direction of R, we note that
B 10 km
tan θ = = = 2
A 5km
With a calculator, we find that the angle whose tangent is closest to 2 is θ = 63 . To express the direction of R in
◦
terms of north, we see from Fig. 2-11 that the angle φ between north and R, plus the angle θ between R and east, is