Force and Newton's Laws

Force and Newton’s Laws

Chapter

Units of the Chapter

• 5. Force And Newton’s Laws.

• 5.1 Newton’s First Law.

• 5.2 Force.

• 5.3 Mass.

• 5.4 Newton’s Second Law.

• 5.5 Newton’s Third Law.

• 5.6 Applications of Newton’s Laws.

We focus our attention on the motion of a particular body. It interacts with the surrounding bodies ( its environment) so that its velocity changes: an acceleration is produced. The central problem of classical mechanics is this:

(1) We are given a body whose characteristics (mass, volume, electric charge, etc.) we know.

(2) We place this body, at a known initial location and with a known initial velocity, in an environment of which we have a complete description.

(3) What is the subsequent motion of the body?

5.1 Newton’s First Law

For centuries the problem of motion and its causes was a central theme of natural philosophy, an early name for what we now call physics.

Before Galileo’s time most philosophers thought that some influence or

"force“ was needed to keep a body moving.

They thought that a body was in its "natural state" when it was at rest. For a body to move in a straight line at constant speed,

For example, they believed that some external agent had to continually propel it; otherwise it would "naturally" stop moving.

An external force is needed to set the body in motion, but no external force is needed to keep a body moving with constant velocity .

This principle was adopted by Newton as the first of his three laws of motion:

Newton’s first law helps us to identify this family of reference frames if we express it as follows:

The tendency of a body to remain at rest or in uniform linear motion is called inertia, and Newton’s first law is often called the law of Inertia.

The reference frames to which it applies are called inertial frames.

5.2 Force

As a standard body we find it convenient to use the

standard kilogram and this body has been assigned, by

definition, a mass m

₀

of exactly 1 kg.

• For an environment that exerts a force, we place the standard body on a horizontal table having negligible friction and we attach a spring to it. We hold the other end of the spring in our hand.

• Now we pull the spring horizontally to the right so that by trial and error we are able to give the standard body a measured constant acceleration a of exactly 1 m/s².

• Therefore, the spring is exerting on the standard kilogram a constant force whose magnitude we call "1 newton" (abbreviated 1 N).

The exerted constant force on the standard kilogram =

= Mass x Acceleration

=

m

₀

x a= 1 kg x 1 m / s

²

=

^1N

L: is the normal spring unextended length. ΔL : is the spring stretched amount.

• Let us arrange to exert a force of 4 N along the x axis and a force of 3 N along the y axis.

• We apply these forces first separately and then simultaneously to the standard body placed, as before, on a horizontal, frictionless surface.

What will be the acceleration of the standard body?

Note that we have two methods of analysis available, which should produce identical results:

1. Find the acceleration produced by each separate force, and add the resultant accelerations vectorially.

2. Add the forces vectorially to a single resultant, and then find the acceleration when that single net force is applied to the body.

5.3 Mass

• Let us attach a spring to our standard body, the standard mass (kilogram) m0= 1 kg, to which we have arbitrarily assigned a mass, and arrange to give it an acceleration a0of, say, 2.00 m/s², using the method of Fig. 5.1b.

• If we have two identical standard bodies (2m0= 2 kg), and the spring stretches by the same amount ΔL, thus a₀ =1.00 m/s².

• If we have two identical standard bodies (3m0= 3 kg), and the spring stretches by the same amount ΔL, thus a₀ =0.667 m/s².

• we conclude from many such experiments that the acceleration produced by a given force is inversely proportional to the mass being accelerated.

• The ratio of the masses of the two bodies is then the same as the inverse ratio of the accelerations given to these bodies by that force,

or

For example

, suppose as above we use a force that gives an acceleration a₀ of 2.00m m/s² to the standard body. We apply the same force (by stretching the spring by the same amount L) to a body of unknown mass m₁, and we measure an acceleration a₁ of, say, 0.50 m/s² and knowing that the standard mass is m0 (1 kilogram) _. Therefore

,

we can then solve for the unknown mass, which gives

(For two different forces)

For example

, we apply a greater force so that the extension of the spring is 1.5 L. We would then find that the standard mass m0, is accelerated to 3.00 m/s² and the unknown mass m₁, is accelerated to 0.75 m/s². We would deduce the unknown mass to be

• We can show, in still another experiment of this type, that if objects of mass m₁ and m₂ are fastened together, they behave mechanically as a single object of mass (m₁+ m₂). In other words, masses add like (and are) scalar quantities.

5.4 Newton’s Second Law

We can now summarize all the previously described experiments and definitions in one equation, the fundamental equation of classical mechanics

• In this equation F is the (vector) sum of all the forces acting on the body, m is the mass of the body, and a is its (vector) acceleration.

• We shall usually refer to F as the resultant force or net force.

• The acceleration of the body is in magnitude directly proportional to the resultant force acting on it and in direction parallel to this force.

• As in the case of all vector equations, we can write this single vector equation as three scalar equations,

• The x , y , and z components of the resultant force ( F

^x

, F

^y

, and F

^z

) to the

x , y , and z components of acceleration ( a

^x

, a

^y

, and a

^z

) for the mass m .

1 Sample Problem

Solution

• In drawing free-body diagrams, it is important always to include all forces that act on the particle

• We can then find the acceleration of the sled from Newton’s second law, or

2 Sample Problem

Solution

• Let us find the (constant) acceleration, using Eq.

The negative sign shows that the student is pushing the sled in the direction of decreasing x.

3 Sample Problem

A crate whose mass m is 360 kg rests on the bed of a truck that is moving at a speed v⁰ of 120km/h, as in Fig. 5.4a. The driver applies the brakes and slows to a speed v of 62km/h in 17 s. What force (assumed constant) acts on the crate during this time? Assume that the crate does not slide on the truck bed.

Solution

We first find the (constant) acceleration of the crate. Solving Eq

• The force on the crate follows from Newton’s second law:

5.5 Newton’s Third Law

• Forces acting on a body result from other bodies that make up its environment.

• Any single force is therefore part of the mutual interaction between two bodies.

• We find by experiment that when one body exerts a force on a second body, the second body always exerts a force on the first.

• Newton’s third law can then be stated in traditional form:

• A more modern version of the third law concerns the mutual force exerted by two bodies on one another:

• The following examples illustrate applications of the third law.

1. An orbiting satellite. Figure 5.6 shows a satellite orbiting the Earth. The only force that acts on it is F^SE, the force exerted on the satellite by the gravitational pull of the Earth. Where is the corresponding reaction force? It is F^ES, the force acting on the Earth owing to the gravitational pull of the satellite.

2. A book resting on a table. Figure 5.7a shows a book resting on a table. The Earth pulls downward on the book with a force F^BE. The book does not accelerate because this force is canceled by an equal and opposite contact force F^BT exerted on the book by the table.

3. Pushing a row of crates. Figure 5.8 shows a worker W pushing two crates, each of which rests on a wheeled cart that can roll with negligible friction.

The worker exerts a force F_1Won crate 1, which in turn pushes back on the worker with a reaction force F_{W 1}. Crate 1 pushes on crate 2 with a force F₂₁, and crate 2 pushes back on crate 1 with a force F₁₂. (Note that the worker exerts no force on crate 2 directly.) To move forward, the worker must push backward against the ground. The worker exerts a force F^GW on

Letting a represent the common acceleration and adding the equations give:

This same equation would result if we considered crates 1 and 2 to be a single object of mass (m₁ +m₂), The net external force acting on the combined object is F_1W.

4. Block hangingfrom a spring. Figure 5.9a shows a block hanging at rest from a spring, the other end of which is fixed to the ceiling. The forces on the= block, shown separately in Fig. 5.9b, are its weight W (acting down) and

the force F exerted by the spring (acting up).

• The reaction force to F (the force exerted on the block by the spring) is the force exerted by the block on the spring.

• To show this force, we illustrate the forces acting on the spring in Fig. 5.9c. These forces include the reaction to F, which we show as a force F0 (= -F) acting downward, the weight w of the spring (usually negligible), and the upward pull P of the ceiling. If the spring is at rest, the net force must be zero: P +w+ F0= 0:

5.6 Applications Of Newton’s Laws

The basic steps in applying Newton’s laws are these:

1. Clearly identify the body that will be analyzed. Sometimes there will be two or more such bodies; each is usually treated independently.

2. Identify the environment that will be exerting forces on the body-surfaces, other objects, the Earth, springs, cords, and so on.

3. Select a suitable inertial (nonaccelerating) reference frame.

4. Pick a convenient coordinate system (in the chosen reference frame), locate the origin, and orient the axes to simplify the problem as much as possible. With suitable care, a different coordinate system can be chosen for each component of a complex problem.

5. Make a free-body diagram, showing each object as a particle and all forces acting on it.

6. Now apply Newton’s second law to each component of force and acceleration.

4 Sample Problem

Figure 5.10a shows a block of mass m = 15.0 kg hung by three strings. What are the tensions in the three strings?

Solution

Figure 5.10b shows the free-body diagram of the knot, which remains at rest under the action of the three forces T^A, T^B and T^C, which are the tensions in the strings.

From Fig. 5.10b we see that

Only the y components enter, and again the acceleration is zero:

5 Sample Problem

A sled of mass m = 7:5 kg is pulled along a frictionless horizontal surface by a cord (Fig. 4.11). A constant force of P = 21:0N is applied to the cord. Analyze the motion if (a) the cord is horizontal and (b) the cord makes an angle of = 15 with the horizontal.

Solution

A block of mass m = 18:0 kg is held in place by a string on a frictionless plane inclined at an angle of 27(see Fig. 5.12a).

(a) Find the tension in the string and the normal force extended on the block by the plane.

(b) Analyze the subsequent motion after the string is cut.

6 Sample Problem

Solution

In the static case there is no acceleration and the forces must sum to zero. The weight is resolved into its x (-mg sin) and y (-mg cos) components, and the force equations are as follows

(b) When the string is cut, the tension disappears from the equations and the block is no longer in equilibrium. Newton’s second law now gives the following:

Cutting the string doesn’t change the motion in the y direction (then block doesn’t jump off the plane!), so a^y = 0 as before and the normal force still equals mg cos , or 157 N. In the x direction

The minus sign shows that the block accelerates in the negative x direction, that is, down the plane

7 Sample Problem

A passenger of mass 72.2 kg is riding in an elevator while standing on a platform scale (Fig. 5.13a). What does the scale read when the elevator cab is (a) descending with constant velocity and (b) ascending with acceleration 3.2 m/s²?

Solution

When a = 0, such that the elevator is at rest or moving with constant velocity, as in part (a), then

Homework

1- A hockey puck having a mass of 0.30 kg slides on the horizontal, frictionless surface of an ice rink. Two forces act on the puck, as shown in Figure 5.5. The force F₁has a magnitude of 5.0 N, and the force F₂has a magnitude of 8.0 N.

Determine both the magnitude and the direction of the puck’s acceleration?

2- A traffic light weighing 125 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37.0° and 53.0° with the horizontal. Find the tension in the three cables?

3- A crate of mass m is placed on a frictionless inclined plane of angle.

(a) Determine the acceleration of the crate after it is released?

(b) Suppose the crate is released from rest at the top of the incline, and the distance from the front edge of the crate to the bottom is d. How long does it take the front edge to reach the bottom, and what is its speed just as it gets there?

4- A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator, as illustrated in Figure 5.14. Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from the weight of the fish?

End of

Chapter (5)