## Problems

Problem 3.1. To hit a ball, a female volleyball player jumps in the vertical direction. If her center of mass rises 0.4 m during the airborne portion of the jump, determine the time duration 2t* during which the player is airborne. Determine the velocity of her center of mass at the instant of takeoff.

Problem 3.2. During a triple axel, a figure skater covers a horizontal distance of 3 m and reaches a maximum height of 0.6 m above the ice. Determine the initial velocity of the skater as she lifts off the ice. Answer: vo = 4.28 ei + 3.43 e2 (m/s), where e2 and e2 are unit vectors in horizontal and vertical directions, respectively.

Problem 3.3. In speed skating, the skater pushes off against the ice as the skate is gliding forward (Fig. P.3.3). The direction of push-off is perpendicular to the gliding direction of the skate. This action results in a sinusoidal trajectory of the center of mass of the skater when skating along the straightaways. A sideways push off the right leg causes a leftward movement of the center of mass and vice versa. During the sideways push-off of the right leg of a 70-kg skater, projection of the push-off force onto the horizontal plane was measured to be 3600 N. The duration of the push-off was 0.32 s. The velocity of the center of mass before the pushoff was in the direction of gliding and the speed was equal to 15 m/s. Determine the speed of the center of mass immediately after the completion of the push-off. (Take ei be the direction of the velocity before the push-off [direction of gliding] and let e2 be the

 (a) (b) C Fe2 > A —ei

Figure P.3.3a,b. Schematic diagram of a speed skater during a sideways push (a). The curved line in (b) that is identified with symbol C represents the path of the center of mass of the skater whereas the straight line along the e1 axis represents the path of the skate A.

Figure P.3.3a,b. Schematic diagram of a speed skater during a sideways push (a). The curved line in (b) that is identified with symbol C represents the path of the center of mass of the skater whereas the straight line along the e1 axis represents the path of the skate A.

Figure P.3.4. A man performing upright row to strengthen his back muscles.

direction of the push-off. Note that e1 and e2 make a right-handed coordinate system in the horizontal plane.) Answer: v = 15.0 e1 + 16.4 e2 (m/s).

Problem 3.4. Using a palms-down grip and hands close together, an athlete holds a barbell at the thighs at 0.7 m above the floor. Then he pulls the barbell up to his chin with an acceleration of 6 m/s2 upward (Fig. P.3.4). Compute the force the ground exerts on the athlete at the initiation of motion. The athlete is 75 kg and the barbell weighs 20 kg. Assume that the center of mass of the athlete remains at 0.8 m above the floor during the exercise. Note that this exercise is called upright row. It is designed to work the back muscle trapezius. Answer: F = 1051 N.

Problem 3.5. A man does chin-ups by pulling himself up toward a bar fixed on a side wall (Fig. P.3.5). Let 6 and \$ be the angles the forearm and the upper arm make with the horizontal axis e1, as shown in Fig. P.3.5. Using the fact that the horizontal distance between the man's two hands (D) does not change during the chin-ups, develop an equation that relates 6 to This equation is called a constraint equation. Answer: D - d = 2 (L cos 6 + L cos \$)

Problem 3.6. A man does chin-ups (Fig. P.3.5). Let 6 and \$ be the angles the forearm and the upper arm make with the horizontal axis e1. At time t = 0, 6 = \$ = 45° and d6/dt = d\$/dt = 0. Assuming also that d2\$/dt2 = 2 rad/s2 at t = 0, determine the force (F) exerted by the holding bar on the fists of the man at the initiation of motion. The man weighs 72 kg. Both the upper arm and the lower arm are 37 cm long; the shoulder width d = 54 cm.

Answer: The trunk of the man moves up and down as a single body. Thus the acceleration of the center of mass of the man in the vertical direction is obtained by taking succesive time derviatives of the vertical distance between the shoulders and the bar:

y = -L (sin 9 + sin v = dy/dt = -L [(dd/dt) cos 9 + (d^/dt) cos a = d2y/dt2 = -L [(d29/dt2) cos 9 + (d2^/dt2) cos 0]

Figure P.3.7a,b. A woman performing leg lifts while holding onto bars.

At t = 0, (de/dt) = (d\$/dt) = 0 and (d2\$/dt2) = 2 rad/s. Compute d2e/dt2 by taking the succesive time derivatives of the constraint equation given as the answer of Problem 3.7: d2e/dt2 = -(d2\$/dt2) sin \$/sin e = -2 rad/s. Thus a = 0 and F = 72 kg • 9.81 m/s2 at t = 0.

Problem 3.7. While holding onto bars, an athlete raises her legs upward (Fig. P.3.7) such that the angle \$ her legs make with the vertical axis varies with time in accordance with the following equation:

What is the differential equation governing the angle of inclination 6 of her upper body with the vertical axis? The woman weighs 50 kg and has a height of 1.9 m. One can represent her as composed of two slender rods of equal length L (L = 0.95 m). The mass of her body is then divided into two (m1, m2) and lumped equally at the center of each rod (m1 = m2 = 25 kg). Hint: This is a mathematically tedious problem. The following equations provide a road map for the solution:

vD/A = d(rD/A)/dt rE/A = rB/A + rE/B = -L (sin 6 e1 + cos 6 e2) + (L/2) (sin \$ e1 - cos \$ e2)

Problem 3.8. A seesaw is released from rest at the position shown in Fig. P.3.8. The weight at B is half the weight at D. Using the conservation of moment of momentum with respect to the fulcrum point A, derive a differential equation that governs the time course of the orientation of the seesaw until D hits the ground. Assume that the seesaw itself is weightless. Masses m and 2m are fixed to the seesaw. Answer: d26/dt2 = -(g/3L) cos 6

Figure P.3.8. A seesaw held at rest is released at time t = 0 with two weights acting on opposite sides. The seesaw itself is assumed

weightless.

## Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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