Chapter

Applications of Newton’s Laws

Blinn College - Physics 1401 - Terry Honan

This chapter discusses the kinematics and dynamics of circular motion, and considers the problem of motion in an accelerated frame of reference.  More applications on the second law will also be presented.

. - Forces of Constraint

Often in mechanics problems the motion of objects is constrained.  A force of constraint is responsible for keeping the constraint; it will take on whatever value is needed to maintain its constraint.  Often forces of constraint satisfy inequalities.  These vague general comments will become clearer as examples are given.  The examples we will discuss here are tension forces, normal forces and static friction forces.

Tension Forces as Constraints

Tension is a force of constraint.  The constraint associated with tension is the condition that the length of the rope stays constant.  In our examples with ropes and two masses, this condition is what related the accelerations of the masses.  In Atwood's machine the string not stretching meant that the two masses moved the same distance, implying the accelerations were the same magnitude.  In the example with two pulleys the mass that moved twice the distance of the other had twice the acceleration.  These conditions eventually lead to a value for the tension forces.

The interactive figures above illustrate tension as a force of constraint.  Both show a weight W hanging from a string or rope.  On the left an applied force F pulls downward on the rope.  The larger F gets the larger T=W+F gets; the tension T will take on whatever value is needed to maintain the constraint.  The figure on the right shows the same hanging weight but now the applied force acts upward.  As the applied force F gets larger T=W-F decreases.  It is clear that if F is larger than the weight W, then the string will lost its tension and the mass will accelerate upward.  To maintain the constraint the tension would need to be negative, but it cannot.  The tension force satisfies a simple inequality

T≥0.

This is equivalent to saying: "You can't push with a rope."  Note that negative tensions are possible for rigid objects, like a stick.

Normal Forces as Constraints

The normal force is also a force of constraint.  The constraint is that there is no motion through the surface.

Consider a block sitting on a floor as in the interactive figures above.  Without an applied force, the normal force is the weight of the book W.  If one pushes down on the book with an applied force F, as in the left figure, then the floor pushes up with a larger normal force of N=W+F; as F gets larger, N gets larger.  If a small force F lifts on the block, as described by the figure on the right, then the normal force N=W-F decreases as F increases.  If the lifting force exceeds the weight the normal force the book will lift; N would have to be negative to maintain the inequality.  A surface can only push away from it

N≥0.

A negative normal force would describe suction or glue.

Because the normal force is a constraint it is always found by solving the second law perpendicular to the surface.

(Subscript[F, net, ⊥])=m (Subscript[a, ⊥])

One does not just insert a value of N into a problem solution; it is something you calculate using the above expression.

Friction Between Surfaces

Recall that forces from surfaces break up into two components: the perpendicular component is the normal force and the parallel component is friction.  There are two types of friction between surfaces, static and kinetic.  Static friction is the case when there is no sliding.  Kinetic friction is when there is sliding.  These are very different things.  Static friction is a force of constraint.  Kinetic friction is a dissipative force.

Static Friction

The interactive figure above shows a horizontal pushing force F acting on a block of weight W.  The force of static friction (Subscript[f, s]) will keep the block from sliding.  It must then act opposite to F and must have the same magnitude, (Subscript[f, s])=F.  As F increases, so does (Subscript[f, s]).   It should be clear that if F is large enough the block will slide.  There is a largest value of the force of static friction (Subscript[f, s]^max).  When the block slides then there is no longer static friction and we get kinetic friction (Subscript[f, k]).

Static friction is a force of constraint; the constraint is that there is no sliding.  No sliding implies that the accelerations of the two surfaces are the same.  Since it is a constraint we always solve for its value by using

(Subscript[F, net, ∥])=m (Subscript[a, ∥]).

There is an upper limit on static friction.  The constraint is that the maximum value is proportional to the normal force.

(Subscript[f, s])≤(Subscript[μ, s]) N

(Subscript[μ, s]) is a dimensionless constant called the coefficient of static friction; it is a property of the two surfaces in contact.  At the critical point between sliding and not the inequality is saturated; that is it become equal: (Subscript[f, s]^max)=(Subscript[μ, s]) N.

We will demonstrate static friction with two examples: one with no acceleration and another with acceleration.

Example . - Stationary Block on an Incline

Solution

(Subscript[F, net, ∥])=m (Subscript[a, ∥])  ⟹  m g sin θ-(Subscript[f, s])=0  ⟹  (Subscript[f, s])=m g sin θ

(Subscript[F, net, ⊥])=m (Subscript[a, ⊥])  ⟹  N-m g cos θ=0  ⟹  N=m g cos θ

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m g sin θ≤(Subscript[μ, s]) m g cos θ  ⟹  tan θ≤(Subscript[μ, s])

(Subscript[μ, s])=tan (Subscript[θ, max])=tan 28 °=0.532

Example . - Crate on Flat Accelerating Surface

Solution

(Subscript[F, net, ∥])=m (Subscript[a, ∥])  ⟹  (Subscript[f, s])=m a

(Subscript[F, net, ⊥])=m (Subscript[a, ⊥])  ⟹  N-m g=0  ⟹  N=m g

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m a≤(Subscript[μ, s]) m g  ⟹  a≤(Subscript[μ, s]) g

a=2.5(QuantityUnit)/(((QuantityUnit))^2), (Subscript[μ, s])=0.30  and g=9.80(QuantityUnit)/(((QuantityUnit))^2)   ⟹   2.5(Overscript[≤, ?])2.94

. - Dissipative Friction

When moving with respect to a surface or medium there is a resistive friction force that opposes the relative motion of the object to the surface or medium.  If (OverHat[v]) is the unit vector in the direction of the relative velocity of the object (to the surface or medium) then the friction force is opposite that direction.

(OverVector[f])=-f (OverHat[v])

The values of f for different cases will be discussed below.

Kinetic Friction

Kinetic friction is the dissipative force when there is sliding between two surfaces.  Its direction opposes the direction of the sliding.  The magnitude of the force of kinetic friction is fixed at the value

(Subscript[f, k])=(Subscript[μ, k]) N.

(Subscript[μ, k]) is the coefficient of kinetic friction.  Like the static friction case, it is a property of the two surfaces.

Example . - Block Sliding on an Incline

Solution

(Subscript[F, net, ⊥])=m (Subscript[a, ⊥])  ⟹  N-m g cos θ=0  ⟹  N=m g cos θ

(Subscript[F, net, ∥])=m (Subscript[a, ∥])  ⟹  m g sin θ+(Subscript[f, k])=m a

m g sin θ+(Subscript[μ, k]) m g cos θ=m a   ⟹   a=g sin θ+(Subscript[μ, k]) g cos θ

Solution

(Subscript[F, net, ⊥])=m (Subscript[a, ⊥])  ⟹  N-m g cos θ=0  ⟹  N=m g cos θ

(Subscript[F, net, ∥])=m (Subscript[a, ∥])  ⟹  m g sin θ-(Subscript[f, k])=m a

a=g sin θ-(Subscript[μ, k]) g cos θ

Friction in a Medium

When an object moves through a fluid, a liquid or a gas, it experiences a friction force with a magnitude that varies with the speed.  This is in contrast to kinetic friction which has a fixed magnitude independent of the speed.  There are two simple ways to model friction in a medium: viscous friction, which is proportional to the speed and quadratic drag, which is proportional to the speed squared.

Viscous Friction

Typically, at low speeds when the fluid flow around the object is laminar (smooth) the friction behaves as viscous friction.

f=b v

v is the speed and the constant b depends on the viscosity of the fluid and the geometry and surface texture of the object.

Quadratic Drag

At speeds higher than where the fluid flow around the object becomes turbulent the friction force becomes quadratic drag.

f=c (v^2)

The constant c depends on the viscosity of the fluid and the geometry and surface texture of the object.

. - Springs and Hooke’s Law

Hooke's law is the force law for a spring.  Define x=0 to be the relaxed position of a spring, the equilibrium position, and let x be the amount the spring is stretched from equilibrium.  Hooke's law states that there is a proportionality between F and x, F∝x.  We can introduce a constant of proportionality k called the spring constant or force constant.

F=k x  (Force on spring)

The spring constant is a property of a particular spring; a stiff spring has a large k and a loose spring has a small k.  In this expression F is the force stretching the spring.  Usually we consider the force of the spring itself when considering Hooke's law.  By Newton's third law this is the negative of the force on the spring and we get

F=-k x  (Force of spring).

Note that when the spring is stretched, x is positive and the force is negative.  When the spring is compressed the force is positive and x is negative.

A force is said to be elastic when it satisfies Hooke's law.  We will see that elasticity is quite common for sufficiently small deformations of almost anything.  What is special about springs is they maintain their elasticity over large deformations.  It should be clear that even a spring will violate Hooke's law when stretched a very large distance.

. - Kinematics of Circular Motion

Uniform Circular Motion - Centripetal Acceleration

We will first consider the case of uniform circular motion, where uniform means that the speed stays constant.  Even with a constant speed there an acceleration, since the direction is changing. To find the acceleration we will first consider a finite time Δ t and the average velocity and average acceleration.  We will then let the time approach zero; the average values will then approach the instantaneous values.

Take (Subscript[(OverVector[r]), i]) and (Subscript[(OverVector[r]), f]) to be the initial and final position vectors at times (Subscript[t, i]) and (Subscript[t, f]) where Δ t=(Subscript[t, f])-(Subscript[t, i]).  The velocities at these times are (Subscript[(OverVector[v]), i]) and (Subscript[(OverVector[v]), f]); since the speed is constant these have the same magnitude.

Take the angle between the two position vectors as θ.  In circular motion the velocity is perpendicular to the position vector; it follows that θ is also the angle between the velocities.  The sides opposite the angle θ in the triangles are

Δ(OverVector[r])=(Subscript[(OverVector[r]), f])-(Subscript[(OverVector[r]), i])=(Subscript[(OverVector[v]), ave]) Δ t  and  Δ(OverVector[v])=(Subscript[(OverVector[v]), f])-(Subscript[(OverVector[v]), i])=(Subscript[(OverVector[a]), ave]) Δ t

Looking at just the magnitudes of these vectors we see a pair of isosceles triangles and since the angle is the same for both, we have similar triangles.

Comparing the ratios of similar sides of similar triangles gives

(((Subscript[a, ave]) Δ t)/((Subscript[v, ave]) Δ t))=(v/r).

This can be rewritten as

(Subscript[a, ave])=((v (Subscript[v, ave]))/r).

As Δ t->0 the average velocity and average acceleration approach their instantaneous values, so the magnitudes of the average quantities approach the instantaneous magnitudes.

(Subscript[v, ave])->v  and  (Subscript[a, ave])->a

With this we can write the value of the instantaneous acceleration as

(Subscript[a, c])=((v^2)/r).

As Δ t->0 it is also true that θ->0.  It follows that the velocity and acceleration vectors are perpendicular.  The direction of the acceleration is toward the center.  We will refer to this direction as centripetal and denote the centripetal acceleration as (Subscript[a, c]).

Many modern texts avoid the use of the term centripetal and have adopted the term radial instead. The radial direction could be outward or inward and the term centripetal is unambiguously toward the center. The reason for avoiding the word centripetal is because students confuse it with centrifugal. The centrifugal force is the outward false force associated with the centripetal acceleration.

Uniform Circular Motion - Period, Speed and Radius

For uniform circular motion we can relate the period,  speed and the radius.  This will also give us an alternative expression for the centripetal acceleration in terms of the period.  The period T is defined as the time needed for each full revolution.

It is a simple matter to relate the period to the radius and speed.  Since the speed  is uniform, is the same as the average speed; this is the distance traveled divided by the total time.  The distance traveled in one period is one circumference.  It follows that

v=((2 π r)/T).

Inserting this into our expression for the centripetal acceleration  (Subscript[a, c])=(v^2)/r gives

(Subscript[a, c])=(((((2 π)/T)))^2) r.

This will prove to be a useful expression.  The student should keep in mind that this only applies to uniform circular motion.

Example . - Centripetal Acceleration at the Equator

Solution

r=(Subscript[R, E])=6.38×(10^6)(QuantityUnit)  and  T=24×3600(QuantityUnit)=8640(QuantityUnit)   ⟹   (Subscript[a, c])=(((((2 π)/T)))^2) r=0.0337(QuantityUnit)/(((QuantityUnit))^2)

General Circular Motion - Tangential Acceleration

Now consider the more general case of circular motion where we allow the speed to change.  Generally, the component of the acceleration perpendicular to the motion is related to the change in direction and the component parallel to the direction of motion is related to the change in speed.

The component of the acceleration in the centripetal direction must be the same as in the uniform case.

(Subscript[a, c])=((v^2)/r).

The tangential direction is the direction of the velocity. If the speed is changing then there is also a tangential component of the acceleration (Subscript[a, t]). If the speed is increasing the tangential component is positive (Subscript[a, t])>0 and if decreasing it is negative (Subscript[a, t])<0.

General Two Dimensional Motion and the Effective Radius

Now we consider the most general problem of motion in a plane. Any trajectory may be approximated by a circle at some position.  We refer to the radius of that circle as the effective radius.  We may choose centripetal and tangential directions as before and apply the above expression for the acceleration to the general two dimensional problem.  If the direction is not changing at some instant, then the effective radius is infinite; the centripetal acceleration then becomes zero.

. - Dynamics of Circular Motion

Newton’s Second Law (Subscript[(OverVector[F]), net])= m (OverVector[a])  is the foundation of classical mechanics and, of course, it must still apply when we have circular motion.  We now have new formulas for the acceleration.  We will now consider examples with circular motion.

Examples with Uniform Circular Motion

Example . - Block on a Rotating Disk

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  (Subscript[f, s])=m (Subscript[a, c])

(Subscript[F, net,ver])=m (Subscript[a, ver])=0  ⟹  N=m g

(Subscript[a, c])=(((((2 π)/T)))^2)r

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m (Subscript[a, c])≤(Subscript[μ, s]) m g  ⟹  (Subscript[a, c])≤(Subscript[μ, s]) g  ⟹  (((((2 π)/T)))^2) r≤(Subscript[μ, s]) g

(((((2 π)/T)))^2) (Subscript[r, max])=(Subscript[μ, s]) g  ⟹  (Subscript[r, max])=(((Subscript[μ, s]) g)/(((2 π/T))^2))=0.434(QuantityUnit)

Example . - Rotating Vertical Cylinder in an Amusement Park

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  N=m (Subscript[a, c])

(Subscript[F, net,ver])=m (Subscript[a, ver])=0  ⟹  (Subscript[f, s])=m g

(Subscript[f, s])≤(Subscript[μ, s]) N  ⟹  m g ≤(Subscript[μ, s]) m (Subscript[a, c])  ⟹  g≤(Subscript[μ, s]) (Subscript[a, c])  ⟹  g≤(Subscript[μ, s]) ((v^2)/R)

g=(Subscript[μ, s]) ((Subscript[v, min]^2)/R)   ⟹  (Subscript[v, min])=(Sqrt[((R g)/(Subscript[μ, s]))])

Nonuniform Examples

Example . - Car at the Bottom of a Trough

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  N-m g=m ((v^2)/R)  ⟹  N=m g+m ((v^2)/R)

Example . - Car at the Top of a Hill

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g-N=m ((v^2)/R)  ⟹  N=m g-m ((v^2)/R)

N≥0  ⟹  v≤(Sqrt[R g])  ⟹  (Subscript[v, max])=(Sqrt[R g])

Example . - Roller-Coaster Loop-to-Loop

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g+N=m ((v^2)/R)  ⟹  N=m ((v^2)/R)-m g

N≥0  ⟹  v≥(Sqrt[R g])  ⟹  (Subscript[v, min])=(Sqrt[R g])

Example . - Ball Swung in Vertical Circle

Solution

(Subscript[F, net, c])=m (Subscript[a, c])  ⟹  m g+T=m ((v^2)/L)  ⟹  T=m ((v^2)/L)-m g

T≥0  ⟹  v≥(Sqrt[L g])  ⟹  (Subscript[v, min])=(Sqrt[L g])