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NEWTON’S LAWS OF MOTION

4

4.1.

IDENTIFY: Consider the vector sum in each case. SET UP: Call the two forces F1 and F2 . Let F1 be to the right. In each case select the direction of F2 such that F = F1 + F2 has the desired magnitude. EXECUTE: (a) For the magnitude of the sum to be the sum of the magnitudes, the forces must be parallel, and the angle between them is zero. The two vectors and their sum are sketched in Figure 4.1a. (b) The forces form the sides of a right isosceles triangle, and the angle between them is 90°. The two vectors and their sum are sketched in Figure 4.1b. (c) For the sum to have zero magnitude, the forces must be antiparallel, and the angle between them is 180°. The two vectors are sketched in Figure 4.1c. EVALUATE: The maximum magnitude of the sum of the two vectors is 2F, as in part (a).

Figure 4.1 4.2. IDENTIFY: We know the magnitudes and directions of three vectors and want to use them to find their components, and then to use the components to find the magnitude and direction of the resultant vector. SET UP: Let F1 = 985 N, F2 = 788 N, and F3 = 411 N. The angles θ that each force makes with the

+ x axis are θ1 = 31°, θ 2 = 122°, and θ3 = 233°. The components of a force vector are Fx = F cosθ and

2 2 Fy = F sin θ , and R = Rx + R y and tan θ =

Ry Rx

.

EXECUTE: (a) F1x = F1 cosθ1 = 844 N, F1 y = F1 sin θ1 = 507 N, F2 x = F2 cosθ 2 = − 418 N,

F2 y = F2 sin θ 2 = 668 N, F3 x = F3 cosθ3 = − 247 N, and F3 y = F3 sin θ3 = − 328 N.

2 2 (b) Rx = F1x + F2 x + F3 x = 179 N; R y = F1 y + F2 y + F3 y = 847 N. R = Rx + R y = 886 N; tan θ =

Ry Rx

so

θ = 78.1°. R and its components are shown in Figure 4.2.

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4-1

4-2

Chapter 4

Figure 4.2...