Figure 3.33VectorAhas magnitude53.0 mand direction20.0 º north of thex-axis. VectorBhas magnitude34.0 mand direction63.0ºnorth of thex-
axis. You can use analytical methods to determine the magnitude and direction ofR.
Strategy
The components ofAandBalong thex- andy-axes represent walking due east and due north to get to the same ending point. Once found,
they are combined to produce the resultant.
Solution
Following the method outlined above, we first find the components ofAandBalong thex- andy-axes. Note thatA= 53.0 m,θA= 20.0º,
B= 34.0 m, andθB= 63.0º. We find thex-components by usingAx=Acosθ, which gives
Ax = AcosθA= (53.0 m)(cos 20.0º) (3.16)
= (53.0 m)(0.940) = 49.8 m
and
Bx = BcosθB= (34.0 m)(cos 63.0º) (3.17)
= (34.0 m)(0.454) = 15.4 m.
Similarly, they-components are found usingAy=AsinθA:
Ay = AsinθA= (53.0 m)(sin 20.0º) (3.18)
= (53.0 m)(0.342) = 18.1 m
and
By = BsinθB= (34.0 m)(sin 63.0 º ) (3.19)
= (34.0 m)(0.891) = 30.3 m.
Thex- andy-components of the resultant are thus
Rx=Ax+Bx= 49.8 m + 15.4 m = 65.2 m (3.20)
and
Ry=Ay+By= 18.1 m+30.3 m = 48.4 m. (3.21)
Now we can find the magnitude of the resultant by using the Pythagorean theorem:
(3.22)R= Rx^2 +R^2 y= (65.2)^2 + (48.4)^2 m
so that
R= 81.2 m. (3.23)
Finally, we find the direction of the resultant:
θ= tan−1(R (3.24)
y/Rx)=+tan
−1(48.4 / 65.2).
Thus,
θ= tan−1(0.742) = 36.6 º. (3.25)
CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 99