Week of Jan 13 and 15 - Forces and Straight-line Motion

objectives: 1) Define the addition of two or more forces.

2) Perform simple calculations involving velocity and acceleration.

3) Explain the motion of objects on which forces act.

1. Introduction: We will look briefly at a few scientific principles that underlie all of Nature and gain some introductory practice in scientific reasoning. We need not go through much detail to gain this insight, but we do need to examine concepts like force, velocity and acceleration, kinetic and potential energy, and light and heat energy. This discussion will center only on motion in a straight line. Motion that is not in a straight line is more complex but still follows the same principles. It is also important to realize that true straight-line motion is virtually impossible to achieve except in the imagination. The reason is that the earth rotates and also moves through space. These are not straight-line motions and introduce small effects, but they are small and can safely be ignored.

2. Force: First, we look at the relationship between force and straight-line motion, ignoring that an object can rotate as well as move laterally. Imagine a frictionless horizontal table on which a puck of mass m is placed. If the puck is placed down carefully with absolutely no sideways nudge, it just sits where it is placed. If, on the other hand, the puck is given a brief push, it is observed to coast along the surface at a constant rate. The puck does not change direction, nor does it slow down or speed up but instead moves at a constant speed. Newton's laws state that when the puck is at rest or coasts at a constant rate in a straight line, there is no net force acting on the puck. No net force implies not that the motion is absent but rather that the motion does not change in magnitude or direction! Not only does the puck on the table not change its motion horizontally, it also does not change vertically.

A force is just a push or a pull acting with a definite strength and direction on an object. The pull of gravity downward, giving the puck weight, is an example of a single force. The force the table exerts upward on the puck is another example of a single force. According to Newton's law stated in the previous paragraph, these two forces, the downward pull of gravity on the puck and the upward push on the puck by the table must exactly cancel each other out because the puck does not change motion vertically. If we ignore the rotation of the earth the two forces have exactly the same strength and are exactly opposite in direction. The two forces described add up to zero, a condition called equilibrium. Forces that are in equilibrium produce "zero net effect" on the motion.

When two or more forces act on a body, they do not necessarily act along a straight line. What then is meant by the net force? The difficulty is that unlike ordinary numbers, forces do not merely add. Force can not be represented by a number alone! The direction as well as the strength of each force must be taken into account. When many forces act simultaneously on an object, they must be added by means of special rules.

First, each force is represented by an arrow pointing in the direction of that force and having a length proportional to the strength of that force. In Figure1, the force F₁ has a strength of 4 lbs, F₂ a strength of 2 lbs and F₃ a strength of 3 lbs,. Since F₂ is half as great as F₁, an arrow of half the length of F₁ is drawn as a representation of F₂. Similarly, F₃ is represented by an intermediate length arrow. Second, the tail of each successive "arrow" F₁, F₂ and F₃ is placed on the head of the previous arrow, as shown. The sum of all the arrows is an arrow pointing from the tail of the first to the head of the last. That resultant arrow represents the single force that is equivalent to the sum of the forces.

As an example, three given forces act horizontally on the puck, as shown in Figure 2. The three forces, F₁, F₂ and F₃, oppose each other but along different directions. It is possible for these three forces to balance each other so that the puck feels no net horizontal force. The rule is that if the three force arrows are added as in Figure 1, and the head of F₃ falls directly on the tail of F₁, then the three forces add to zero. If the puck remains on the surface, there is also balance in the vertical sense. The downward pull of gravity is balanced by the upward force the surface exerts on the puck from below. The mass m would then behave as if there were no force at all. There is a mathematical approach, called vector algebra, to determine the net effect of a combination of forces, but we will not deal with this, focusing instead on the visual appearance of balance and on understanding what a balance of forces means conceptually.

3. Velocity and Acceleration in a Straight Line. We have learned that an object at rest or in straight-line motion at constant speed must have all forces acting on it add to zero. But what about the effects on an object when the forces acting are not in balance? For this Newton’s second law is needed. For any object in straight-linemotion, there are two useful quantities we can use to describe its motion. We can state its velocity and acceleration as a function of time. The definitions of velocity and acceleration are as follows.

1. If an object moves in a straight line, the velocity of the object is defined by choosing a small interval of time and noticing how far the object moves during that time interval. An object that moves 2.0 meters along a straight path for 0.30 seconds has an average velocity of

v = (2.0 meters)/(0.30 seconds) = 6.7 m/s in that direction.

2. Acceleration is a less familiar concept. The acceleration of an object is defined by choosing a small time interval again but now noting how much the velocity changes during that time interval. If our object is moving at 2.0 m/s, and 0.5 seconds later is moving 2.3 m/s in the same direction, the average acceleration is

a = (2.3 m/s 2.0 m/s)/(0.5 s) = 0.6 m/s².

In other words, the velocity increases at the rate 0.6 m/s each second. A velocity or an acceleration can be positive or negative. An object with a positive velocity and a negative acceleration is slowing. An object with a negative velocity and a negative acceleration is speeding up in the negative direction.

4. Velocity and Acceleration for Curved Motion: If the path is not straight, the concepts of velocity and acceleration are more complex. The added complexity is that, just like force, neither velocity nor acceleration can be described by a number alone. Direction must be taken into account. Quantities like force, velocity and acceleration must all be treated mathematically as vectors. When an object moves along a curved path, part of the acceleration is along the path and part is perpendicular to the path. A useful special case occurs when an object moves in a circle at a constant speed. Then, the acceleration is completely perpendicular to the path and points toward the center of the circle. This perpendicular acceleration has the name "centripetal acceleration," and for the special case of circular motion at a constant speed always has the value

a(centripetal) = v²/R.

5. Newton's Laws of Motion: Of the two quantities, velocity and acceleration, acceleration is the one most closely connected with the basic laws of physics known as Newton's first and second laws. As stated earlier, these laws state that when all the forces acting on an object of mass m are balanced, the mass has zero acceleration. They also state that when all the forces are not balanced, the acceleration of the mass is found by dividing the NET effective force by the mass

a = F/m.

The direction of the acceleration is identical to the direction of the net force. For calculations, the mass unit is the kilogram (kg), and the force unit is the Newton (N).

4. Example: What happens when a tennis ball of mass m is dropped from great height? At the moment the ball is released, it starts to fall because of the unbalanced pull gravity exerted downward. The pull of gravity is a force that is a determined by the object's mass, and this force is called the weight of the object. The larger the mass, the larger the weight. A 1-kg mass weighs 9.8 N. Measurements show that the dropped ball (as well as a brick or an elephant) accelerates initially at a value very close to 9.8 m/s², a value called "g." One tenth second after release, the ball will have acquired a velocity of 0.98 m/s; 0.2 seconds after release, the ball will have acquired a velocity of 1.96 m/s, etc. However, as the ball's speed increases, wind drag starts to exert a noticeable retarding force upward. Examine Figure 3 showing a falling tennis ball. The gravity force and the wind drag oppose each other. But the wind drag force grows larger and larger as the ball moves faster and faster, much as the force of the wind against your face increases as you speed up on your motorcycle. As the drag force on the dropped ball grows, the net force therefore decreases. Since the net force decreases, the acceleration also decreases, and the ball increases in velocity more slowly. Within a few seconds, when the ball has reached a velocity of perhaps 30 or 40 m/s, the wind drag force grows large enough such that the upward drag force balances the downward gravity force. The acceleration tends toward zero, and the ball no longer increases its velocity. The ball is said to be falling at its terminal velocity. The terminal velocity of a falling object is determined by the behavior of the wind drag. A parachute is designed to provide enough of a drag effect that a person's terminal velocity is small.

Answers to assigned questions in Chapter 3

Ex. 2. His speed relative to the shore will be zero

Ex. 14. Yes to both questions. An object sliding on a horizontal frictionless table has zero acceleration. When a ball is tossed upward, its acceleration is always downward at 9.8 m/s², even at the top of the trajectory, when the velocity is instantaneously at rest.

Ex. 26. In this question, air resistance is assumed to be zero. Whether going up or down, the acceleration is 9.8 m/s² downward. On the upward part of its flight, the ball slows down 9.8 m/s each. On the downward part of its flight, the ball speeds up 9.8 m/s each second. The time going up and the time coming down are exactly the same.

Ex. 38. As water falls from the faucet, it accelerates downward at 9.8 m/s². Thus, the stream has water which is moving faster, the farther it is from the faucet. A higher velocity demands a thinner stream. Eventually, the thinning stream breaks up into droplets.

Prob. 2. The acceleration is the rate of change of velocity. A velocity change of 100 km/hr occurring in a time interval of 10 seconds gives an acceleration of (100 km/hr)/(10 s) = –10 km/hr^.s. Since there are 3600 seconds in an hour, and 1000 meters in a kilometer, this acceleration can also be epressed as

Prob. 3. We will use 10 m/s² as the acceleration due to gravity, and we will define up as +. The initial velocity is +30 m/s. Since the ball loses 10 m/s each second, the time going up is 3 seconds. The average velocity going up is +15 m/s, so that the height at the peak is (3 s)(+15 m/s) = 45 m. The time for the ball to fall back is also 3 seconds, which means that the total time of flight is 6 s.

Prob. 7. This often is a misunderstood question. One can not just average the two speeds. Instead, one must find the total distance and divide by the total time. At a speed of 200 km/hr, the plane takes 3 hours to fly 600 km. The trip back, going at a speed of 300 km/hr, takes only 2 hours. Therefore, the total distance of 1200 km is done in a total time of 5 hours for an average speed of 240 km/hr.

Prob. 8. This is just like the previous problem, except that the distance is not given. This means that we should just assume a distance, call it d, and hope for the best. As it turns out the distance is not relevant because it will drop out. Let us call the time going t₁ and the time coming back t₂. In the calculation, I will suppress the units and just remember that the speed is in km/hr. Then

t₁ = d/v = d/40

t₂ = d/60

The total time is the sum t₁ + t₂ = d/40 + d/60 = 5d/120. The total distance is 2d. Therefore, the average speed is

v(average) = (distance)/(total time) = (2d)/(5d/120) = 48 km/hr.

Incidently, now that we have discovered the distance is idependent of the answer, you can always just make up a convenient number for the distance ans use it!

Answers to assigned questions in Chapter 4

Ex. 8. When you toss a ball upward, the net upward force must be upward. This means the upward force of your hand on the ball must be greater than than the downward force of gravity (its weight).

Ex. 10. The net force on the crate must be zero since the crate remains at rest. The friction force must therefore be equal and opposite to the applied forde F. The friction force is –F.

Ex. 24. Again, the net force on the wagon must be zero, since the wagon is at constant velocity. Thus, there must be a friction force that balances the pulling force.

Ex. 25. Note that 30 N pulls three blocks and produces an acceleration a. All three blocks are accelerating at the same rate. Therefore, only 20 N is needed to pull 2 blocks with the samr acceleration a. This also implies the tension in the rope between the two right hand blocks. For the same reason, 10 N is the tension between the left-hand and middle block. If friction were present, it would have to be taken into account for each block.

Ex. 37. We discussed this at length in class. The downward force of gravity and the upward force of air drag must be equal.

Prob. 2. Newton’s second law states that the net force = mass times acceleration. Then, the second law gives us the equation 200 N = (40 kg)(a). Solving for a leads to the result a = 5 m/s².

Prob. 6. I hope it is clear to you that in the absence of friction, the answer must be that the larger unknown mass M = 4m. The formal way of seeing this is to set up a ratio. Since the acceleration for each case is the same, and since Newton’s second law is a = F/m, then F/m = 4F/M. M must be 4m.

Prob. 9. Acceleration is defined as a = (change in velocity)/(time interval). Thus, the runner accelerates at

a = (1 m/s)/(2 s) = 0.5 m/s². The force of the ground acting on him to cause this acceleration must be according to Newton’s second law, or F = ma. We find F = (60 kg)(0.5 m/s²) = 30 N.

Answers to assigned questions in Chapter 3

Answers to assigned questions in Chapter 4

More Questions for Thought