## Summary

Newtons's laws describe the interaction between forces and motion. Newton's first law states that a particle will remain at rest unless it is acted on by an unbalanced (resultant) force. Newton's second law is about how forces acting on a particle affect its motion:

2F = m a where 2F denotes the resultant force acting on a particle, m is the mass of the particle, and a is its acceleration, measured with respect to a coordinate system fixed on earth. According to this law, a particle will accelerate in the direction of the resultant force acting on the particle. The magnitude of the acceleration will be equal to the magnitude of the resultant force divided by the mass of the particle. Newton's first law is merely a subset of the second law. A particle is an object, large or small, in which the variations in velocity on its body is negligibly small at any instant when compared to the mean velocity of the object at the same instant. A particle could be a star, a baseball, or even an ice-skater gliding on ice.

Newton's third law describes how objects, living as well as nonliving, interact with each other:

in which f1-2 is the force exerted on particle 1 by particle 2. This law states that reactive force to an action is equal in magnitude to the force of action but is in the reverse direction. Contrary to common belief, when a bull hits a bullfighter and sends him off into the air, the fighter actually hits the bull with the same intensity; it is just that the bull has a much larger mass and therefore can easily withstand the force.

### 2.7 Problems

Problem 2.1. Consider a cable connecting two points A and B that are fixed in space (Fig. P.2.1). The cable is in tension so that it forms a straight line passing through A and B. Why is the tension in the cable uniform between these points? Take a small segment of the cable and draw all the forces that act on it. Use Newton's second law for this segment of the cable.

Problem 2.2. Why does the tension in an inextensible cable remain uniform when the cable goes over a frictionless pulley? Problem 2.3. The coordinates of a point with respect to a Cartesian coordinate system E with unit vectors (e1, e2, and e3) can be written as A (x1, x2, x3) where A denotes the point under question. Let A (1, 5, -2) and B (3, -4, 6) be two points in space. Determine the position vector connecting A to B. Also determine the unit vector along the straight line from A to B.

Answer: rB/A = 2 e1 - 9 e2 + 8 e3; e = 0.16 e1 - 0.74 e2 + 0.66 e3.

Problem 2.4. Determine the velocity and acceleration of a small child in a swing of length 3.5 m at a time when the swing is at 30° with the ver-

Figure P.2.1. Two men pulling on a rope.

tical axis. At that instant, the swing was moving up at a constant rate of (^/8) rad/s.

Answer: v = 1.2 ei + 0.7 e2 (m/s); a = 0.27ei + 0.47e2 (m/s2) where ei and e1 are unit vectors in the plane of motion in the horizontal and vertical directions, respectively.

Problem 2.5. A volleyball, thrown from position r = 2 e2 at time t = 0, occupies the position r = 6 e1 + 4 e2 at time t = 2 s. Determine the initial velocity of the ball. Determine the highest elevation (y) the ball reaches while airborne. Answer: vo = 3 e1 + 10.8 e2 (m/s); y = 7.9 m.

Problem 2.6. In baseball, a pitcher standing 60 ft from the home plate needs about 0.4 s to raise his glove in defense of a hard line drive hit at him. In March 1996, a batter hit the pitcher in the face with a ball that traveled back in 0.29 s. Five of the pitcher's teeth were shattered, and he needed 65 stitches to mend his face. Determine the horizontal component of the velocity of the baseball as it hit the pitcher. Answer: v1 = 207 ft/s.

Problem 2.7. A person drags a sulking dog (who is pretending to be dead) on a polished wood floor along a straight line at a rate of 0.50 m/s. The angle the taut leash makes with the horizontal is 30°. The dog weighs 20 kg and the coefficient of dynamic friction f = 0.2. Determine the tensile force exerted on the dog by the leash. What force does the dog exert on her 'inconsiderate' owner? Note that the frictional force acting on the dog is equal to the contact force normal to the floor times the coefficient of dynamic friction. Answer: T = 40.6 N.

Problem 2.8. A person drags a sulking dog on a polished wood floor along a straight line at an acceleration of 0.40 m/s2. The dog pretends to be dead as if she were incapable of moving her limbs. The angle the taut leash makes with the horizontal is 60°. The dog weighs 20 kg and the coefficient of dynamic friction f = 0.2. Determine the tensile force exerted on the dog by the leash. Answer: T = 70.2 N.

Problem 2.9. A woman doing seated machine curls rests her upper arm on an arm pad, parallel to the floor as shown in Fig. 2.12. She then pulls on the bar that is connected through a cable and a pulley to a weight of 10 kg. The weight moves up with an acceleration of 2 m/s2. Determine the tension in the cable. Answer: T = 118 N.

Problem 2.10. A woman performs seated machine curls as shown in Fig. 2.12. She performs the exercise slowly so that the acceleration of the weights she lifts can be neglected. If she lifts a weight of 15 kg, deter mine the force exerted by the bar on the woman when her forearm makes 30° with the horizontal axis. Hint: first determine the unit vector along line BD.

Problem 2.11. Experiments on a wind tunnel showed that the frictional force exerted by air on an object can be estimated by the following formula:

where p is the mass density of air (p = 1.2 kg/m3), A is the area of projection of the object normal to the direction of motion, Cd is the di-mensionless drag coefficient (Cd = 0.8), and u is the speed of the moving object. Provide an upper-bound estimate for the maximum resistance force exerted by air on a 5-kg steel ball falling from a distance of 60 m. How big is the air resistance in comparison with the force of gravity? The diameter of the ball is 6 cm. Hint: Determine the velocity of the ball at y = 0 m by neglecting the effect of air resistance on the fall, and then use this velocity to compute the resistance force. Answer: F = 1.6 N, W = 49.1 N.

Problem 2.12. Determine the arm length and the period of swing of 15 subjects (students in a classroom) and see whether the data are consistent with the predicted dependence of the period of the pendulum on the length of the pendulum (Eqn. 2.18e).

Problem 2.14. Provide a proof for the velocity and acceleration expressions in the path coordinates. The proof can be found in most books on rigid body dynamics.