Newton’s Laws Applications

Newton’s Laws

Applications

Introduction to Friction

Introduction to Friction

Friction Friction is the force that opposes a sliding

motion. Friction is due to microscopic irregularities in

even the smoothest of surfaces. Friction is highly useful. It enables us to walk

and drive a car, among other things. Friction is also dissipative. That means it

causes mechanical energy to be converted to heat. We’ll learn more about that later.

Microscopic View

W

N

Friction may or may not exist between two surfaces. The direction of friction, if it exists, is opposite to the direction object will slide when subjected to a horizontal force. It is always parallel to the surface.

Fpushf(friction)

Small view:Microscopic irregularities resist movement.

Big view:Surfaces look perfectly smooth.

Friction depends on the normal force.

The friction that exists between two surfaces is directly proportional to the normal force.

Increasing the normal force increases friction; decreasing the normal force decreases friction.

This has several implications, such as… Friction on a sloping surface is less than friction on a

flat surface (since the normal force is less on a slope). Increasing weight of an object increases the friction

between the object and the surface it is resting on. Weighting down a car over the drive wheels increases

the friction between the drive wheels and the road (which increases the car’s ability to accelerate).

Static Friction This type of friction occurs

between two surfaces that are not slipping relative to each other.

fs sN fs : static frictional force (N) s: coefficient of static friction N: normal force (N)

fs < sN is an inequality! The fact that the static friction equation is an

inequality has important implications. Static friction between two surfaces is zero unless

there is a force trying to make the surfaces slide on one another.

Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by s.

Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surface

fs = 0There is no static friction since there is no applied horizontal force trying to slide the book on the surface.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfs

0 < fs < sN and fs = FStatic friction is equal to the applied horizontal force, and there is no movement of the book since F = 0.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfs

fs = sN and fs = FStatic friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfk

fs = sN and fs < FStatic friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force.

Static friction on a ramp

PhysicsN

surface

f s

Wx = mgsin and N = mgcosAt maximum angle before the book slides, we can prove that s = tan

W = mg

Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there.

Static friction on a ramp

PhysicsN

surface

f s

fs = mgsin and N = mgcosAt maximum angle before the book slides, we can prove that s = tan

W = mg

F = 0Wx = fsmgsin =smgcoss = sincos = tan

Assume is maximum angle for which book stays in place.

x

W x

Kinetic Friction This type of friction occurs between

surfaces that are slipping past each other. fk = kN

fk : kinetic frictional force (N) k: coefficient of kinetic friction N: normal force (N)

Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.

Sample ProblemA 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30.a) What is the maximum horizontal force that can be applied to the box before it begins to slide?b) What force is necessary to keep the box sliding at constant velocity?

Sample ProblemA 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp ifa) no force horizontal force is applied to the box?b) a 20 N horizontal force is applied to the box?c) a 60 N horizontal force is applied to the box?

ProblemA 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp ifa) the ramp is at a 25o angle?b) the ramp is at a 45o angle?c) what is the acceleration of the box when the ramp is at 45o?

Strings and Springs

Tension Tension is a pulling force that arises

when a rope, string, or other long thin material resists being pulled apart without stretching significantly.

Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

A physical picture of tension

Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string.Copyright James Walker, “Physics”, 1st ed.

Tension examples

Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force. Copyright James Walker, “Physics”, 1st ed.

Sample problemA. A 1,500 kg crate hangs motionless from a crane cable.

What is the tension in the cable? Ignore the mass of the cable.

B. Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now?

Springs (Hooke’s Law) The magnitude of the force exerted by a

spring is proportional to the amount it is stretched.

F = kx F: force exerted by the spring (N) k: force constant of the spring (N/m or N/cm) x: displacement from equilibrium (unstretched

and uncompressed) position (m or cm) The direction of the force is back toward

the equilibrium (or unstretched) position.

Sample problem• A 1.50 kg object hangs motionless from a spring with a

force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

Sample problem• A 1.80 kg object is connected to a spring of force constant 120

N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60.

Copyright James Walker, “Physics”, 1st ed.

Connected Objects

Sample problemA 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown,A) what is the acceleration of the system?B) what is the tension in the string connecting the objects?(Assume a frictionless surface.)

Copyright James Walker, “Physics”, 1st ed.

Gravity

A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.

Slowing gravity down

Magic pulleys on a flat table

Magic pulleys bend the line of action of the force without affecting tension.

Frictionless tablem1

m2

T

m2g

N

m1g

T

-x

x

F = mam2g + T – T = (m1 + m2)aa = m2g/(m1+m2)

Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find(a)the acceleration of each block. (b)the tension in the connecting string.

Sample problem

m1

m2

Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary.

Sample problem

m1

m2

Sample problem - solution

m1

m2

T

m2g

N

m1g

Tfs

F = 0m2g - T + T – fs = 0fs = m2gsN = m2gsm1g = m2gs = m2/m1 = 0.50

Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.30 and k = 0.20, what is (a)the acceleration of each block? (b)the tension in the connecting string?

Sample problem

m1

m2

Note: we know from previous problem that the static friction is not enough to hold the blocks in place!

Sample problem – solution (a)

m1

m2

T

m2g

N

m1g

T

F = mam2g - T + T – fk = ma

m2g - km1g = (m1 + m2)aa = (m2- km1) g/(m1 + m2)

a = 2.0 m/s2fk

Sample problem – solution (b)

m1

m2

T

m2g

N

m1g

T

Using block 2F = ma

m2g - T = m2aT = m2(g – a)

T = 40 Nfk

Using block 1F = ma

T - fk = m1aT = m1(a + kg)

T = 40 N

Using that the acceleration is 2.0 m/s2 from part a)

Pulleys and Ramps - together

Ramps and Pulleys – together!

Magic pulleys on a ramp It’s a little more complicated when a magic pulley is

installed on a ramp.

m 1 m2

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)

m1g

N T

T

m2g

m1gsin

m1gcos

Sample problem Two blocks are connected by a string as

shown in the figure. What is the acceleration, assuming there is no friction?

10 kg5 kg

Sample problem - solution

10 kg5 kg

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)a = [(5 – 10sin45o)(9.8)]/15a = - 1.35 m/s2

m1g

N T

T

m2g

m1gsin

m1gcos

Sample problem - solution

10 kg5 kg

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)a = [(5 – 10sin45o)(9.8)]/15a = - 1.35 m/s2

m1g

N T

T

m2g

m1gsin

m1gcos

How would this change if there is friction on the ramp?

Uniform Circular Motion

Uniform Circular Motion

Uniform Circular Motion An object that moves at uniform speed

in a circle of constant radius is said to be in uniform circular motion.

Question: Why is uniform circular motion accelerated motion?

Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction.

Centrifugal Force Question: What is centrifugal force? Answer: That’s easy. Centrifugal force

is the force that flings an object in circular motion outward. Right?

Wrong! Centrifugal force is a myth! There is no outward directed force in

circular motion. To explain why this is the case, let’s review Newton’s 1st Law.

Newton’s 1st Law and cars•When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward.

• You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward.

•When a car brakes suddenly, you as a passenger feel as if you are flung forward.

• You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates.

You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”.

You are NOT flung to the outside. Your inertia resists the inward acceleration and your body simply wants to keep moving in straight line motion!

As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward.

When a car turns

Centripetal Acceleration Centripetal (or center-seeking)

acceleration points toward the center of the circle and keeps an object moving in circular motion.

This type of acceleration is at right angles to the velocity.

This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object.

Centripetal Acceleration ac = v2/r

ac: centripetal acceleration in m/s2

v: tangential speed in m/s

r: radius in meters

v ac

Centripetal acceleration always points toward center of circle!

Centripetal Force A force responsible for

centripetal acceleration is referred to as a centripetal force.

Centripetal force is simply mass times centripetal acceleration.

Fc = m ac Fc = m v2 / r

Fc: centripetal force in N v: tangential speed in m/s r: radius in meters

Fc

Always toward center of circle!

Any force can be centripetal The name “centripetal” can be applied

to any force in situations when that force is causing an object to move in a circle.

You can identify the real force or combination of forces which are causing the centripetal acceleration.

Any kind of force can act as a centripetal force.

Static frictionAs a car makes a turn on a flat road, what is the real identity of the centripetal force?

Tension

As a weight is tied to a string and spun in a circle, what is the real identity of the centripetal force?

GravityAs the moon orbits the Earth, what is the real identity of the centripetal force?

Normal force with help from static friction

As a racecar turns on a banked curve on a racing track, what is the real identity of the centripetal force?

Tension,with some help from

gravityAs you swing a mace in a vertical circle, what is the true identity of the centripetal force?

Gravity, with some help from

the normal forceWhen you are riding

the Tennessee Tornado at

Dollywood, what is the real identity of

the centripetal force when you are on a vertical loop?

Sample problem• A 1200-kg car rounds a corner of radius r = 45 m. If the

coefficient of static friction between tires and the road is 0.93 and the coefficient of kinetic friction between tires and the road is 0.75, what is the maximum velocity the car can have without skidding?

Sample problemYou whirl a 2.0 kg stone in a horizontal circle about your head. The rope attached to the stone is 1.5 m long.

a) What is the tension in the rope? (The rope makes a 10o angle with the horizontal).

b) How fast is the stone moving?