## Ar

3=60

Figure 2.11a,b. Parabolic trajectory of a golf ball in air. The circles indicate the position of a golf ball at various time intervals. The direction of initial velocity is the same for the trajectories shown in (a). Figure 2.11b illustrates trajectories with the same initial speed.

Integrating these equations with respect to time we find v1 = v10; v2 = -gt + v20; v3 = v30 (2.20a)

X1 = v10 t + X10; X2 = —gt2/2 + v20 t + X20; X3 = v30 t + X30 (2.20b)

The symbols v10 and x10 refer to components of velocity and position at t = 0. To determine the trajectory of the golf ball completely, we need to specify these constants. We assume the ball is at the origin of the reference frame E and its initial velocity is in the (X1,X2) plane:

X10 = X20 = X30 = 0; v10 = Vo cos a, v20 = Vo sin a, v30 = 0 at t = 0

where Vo is the speed of the ball at t = 0 and a is the angle that the initial velocity makes with the e1 axis. Then, for all latter times, the following equations determine the trajectory traversed by the golf ball:

Hence, the golf ball moves with constant speed along the e1 direction, whereas its velocity decreases at a constant rate in the e2 direction and the golf ball has no velocity in the e3 direction. The velocity of the golf ball in the e2 direction becomes equal to zero when the ball reaches the maximal height h, and that occurs at t* = (Vo/g) sin a. The maximum height reached by the golf ball (h) and the horizontal distance traveled (s) are given by the following equations:

From Eqn. 2.21, it is clear that the initial velocity of the golf ball determines the height it reaches and the horizontal distance it covers during free fall. For example, if the initial velocity is 100 m/s and a = 30°, the ball climbs as high as 12.5 m in 5 s and covers a horizontal distance of 86 m before it touches the ground. The centroid of the ball traverses a parabola, and this trajectory is independent of the size or shape of the particle so long as the air friction is negligible in comparison to the force of gravity.

Example 2.10. Machine Curls and the Biceps Muscle. A woman performs seated machine curls to strengthen her biceps muscles (Fig. 2.12). She uses a weight of 10 kg. Horizontal distance between her elbows (A) and the bottom of the frictionless pulley (B) is 60 cm. Vertical distance between the same points is 20 cm. Determine the force exerted by the holding bar on the woman's hands when her arm is at 45° with the horizontal. At that moment, the weight has an upward acceleration of 3 m/s2.

Solution: Let us first consider the forces exerted on the 10-kg weight. Tension T in the cable pulls it up whereas the gravity pulls it down, and Newton's second law dictates that  Figure 2.12. A woman performs seated machine curls to strengthen her biceps muscles. The points A, B, and D denote, respectively, her elbow, the lowest point of the pulley, and the holding bar. The pulley is frictionless.

The tension T in the cable is uniform since the pulley is frictionless.

Next, let us consider the forces acting on the holding bar connected to the cable (see point D in Fig. 2.12). If we neglect the weight of this lightly weighted bar, we find that the tension in the cable must be equal in magnitude but opposite in direction to the force exerted by the woman on the bar. Newton's third law then dictates that the tension in the cable is equal to the force exerted by the bar on the hands of the woman holding the bar.

Now that we know the magnitude of the force, we need to determine its direction. To do that we need to compute the orientation of the unit vector in the direction of the cable connecting points B and D. Using the parallelogram law we can show that rD/B = rA/B + rD/A

in which rD/B denotes the position vector from B to D. All other position vectors in the equation follow the same notation. With respect to the coordinate system E shown in the figure, it can be shown that rA/B = 0.6 e1 - 0.2 e2

rD/B = (0.6 - 0.32 X cos 45°) e1 + (-0.2 + 0.32 X sin 45°) e2 = 0.37 e1 + 0.03 e2

The unit vector eD/B along the position vector rD/B can be found by using Eqns. 2.2b and 2.2c:

eD/B = (0.37 e1 + 0.03 e2)/(0.372 + 0.032)1/2 = 0.99 e1 + 0.08 e2 Thus the force exerted by the woman on the holder is equal to F = T eD/B = 128.1 (0.99 e1 + 0.08 e2) N

Note that this force is approximately horizontal. If the woman were to perform arm curls with free weights, the force exerted on her hands would always be directed vertical downward. 