Page 41 - Schaum's Outline of Theory and Problems of Applied Physics

26 VECTORS [CHAP. 2

The angle between the force each tugboat exerts on the ship and the direction of the ship’s motion is 30 . Hence,
◦
each force has a component in the direction of the ship’s motion of
F x = F cos θ = (6 tons)(cos 30 ) = 5.2 tons
◦
Since there are two tugboats, the resultant force on the ship is
R = 2F x = (2)(5.2 tons) = 10.4 tons

SOLVED PROBLEM 2.14
A boat is headed north at a velocity of 8.0 km/h. A strong wind is blowing whose pressure on the boat’s
superstructure causes it to move sideways to the west at a velocity of 2.0 km/h. There is also a tidal current
◦
present that flows in a direction 30 south of east at a velocity of 5.0 km/h. What is the boat’s velocity
relative to the earth’s surface?
The first step is to establish a suitable set of coordinate axes, such as shown in Fig. 2-17(a). Next we draw the
three velocity vectors A, B, and C and calculate the magnitudes of their x and y components. We find these values:

Fig. 2-17

x components y components
A x = 0 A y = 8.0 km/h
B x =−2.0 km/h B y = 0
C x = C cos 30 ◦ C y =−C sin 30 ◦
= (5.0 km/h)(0.866) =−(5.0 km/h)(0.500)
= 4.3 km/h =−2.5 km/h

These components are shown in Fig. 2-17(b).