Lecture Outlines Chapter 6 Physics, 3

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Lecture Outlines Chapter 6

Physics, 3^rd Edition James S. Walker

Chapter 6

Applications of Newton’s

Laws

Units of Chapter 6

• Frictional Forces

• Strings and Springs

• Translational Equilibrium

• Connected Objects

• Circular Motion

6-1 Frictional Forces

Friction has its basis in surfaces that are not completely smooth:

6-1 Frictional Forces

Kinetic friction: the friction experienced by surfaces sliding against one another

The static frictional force depends on the normal force:

(6-1)

The constant is called the coefficient of kinetic friction.

6-1 Frictional Forces

6-1 Frictional Forces

The kinetic frictional force is also independent of the relative speed of the surfaces, and of their

area of contact.

Example 6-1

Pass the Salt – Please

Example 6-2 (p.145) (x=3,0 m, θ=23º, μ_k= 0,26, t =?) Making a Big Splash

The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum,

6-1 Frictional Forces

depending on what is needed to keep the sum of forces zero.

6-1 Frictional Forces

(6-2) where

(6-3)

The static frictional force is also independent of the area of contact and the relative speed of the surfaces.

Example 6-3 (p.147) (m=95,0 kg, θ = 23,2 º, μ_s=?) Slightly Tilted

Conceptual Checkpoint 6-1 Friction for Rolling Tires

Figure 6-4

Stopping distance with and without ABS

6-2 Strings and Springs

When you pull on a string or rope, it becomes taut. We say that there is tension in the string.

6-2 Strings and Springs

The tension in a real rope will vary along its length, due to the weight of the rope.

Here, we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated.

6-2 Strings and Springs

An ideal pulley is one that simply changes the direction of the tension:

Example 6-4 (p.151) (T=165 N, m=?)

A Bad Break: Setting a Broken Leg with Traction

Conceptual Checkpoint 6-2

Compare the Readings on the Scales

6-2 Strings and Springs

Hooke’s law for springs states that the force increases with the amount the

spring is stretched or compressed:

[Konstanten k kallas fjäderkonstanten och vi kommer bara att behandla ideala fjädrar, det vill säga masslösa fjädrar som lyder

Hookes lag]

Figure 6-8 Spring forces

Active Example 6-2 (p.153)

(F=0,22N, x=3,5 mm. a) k = ? b) Om x = 4,0 mm, F =? ) Nasal Strips

6-3 Translational Equilibrium

When an object is in translational equilibrium, the net force on it is zero:

(6-5)

This allows the calculation of unknown forces.

6-3 Translational Equilibrium

Conceptual Checkpoint 6-3 Comparing Tensions

Conceptual Checkpoint 6-3 Comparing Tensions

Example 6-5 (p.156) (m=6,20 kg, θ=40,0º, T₁=?, T₂=?) Suspended Vegetation

Figure 6-16

Conceptual Exercise 1

Active Example 6-3 (p.157) (m = 1,94 kg, θ = 3,50º, T = ?) The Forces in a Low-Tech Laundry

Chapter 6 Opener Newton’s Laws

6-4 Connected Objects

When forces are exerted on connected objects, their accelerations are the same.

If there are two objects connected by a string, and we know the force and the masses, we can find the acceleration and the tension:

6-4 Connected Objects

We treat each box as a separate system:

6-4 Connected Objects

F –T = m₁a₁ = m₁a

T = m₂a₂ = m₂a (6-7)

F = m₁a + m₂a = (m₁+m₂)a

a = F/(m₁+m₂) (6-8) T = m₂ F/(m₁+m₂) (6-9)

6-4 Connected Objects

If there is a pulley, it is easiest to have the

coordinate system follow the string (NO FRICTION)

Conceptual Checkpoint

T = m₁a

W₂ - T = m₂a W₂ = m₂g

W₂ = (m₁+m₂)a

a = m₂g/(m₁+m₂)

T = m₁m₂g/(m₁+m₂)

Example 6-7 (p.161) Atwood’s Machine

Example 6-7 Atwood’s Machine

T - m₁g = m₁a

m₂g - T = m₂a Eliminate T by adding the two equations

g(m₂- m₁) = a(m₂+m₁)

a = (m₂-m₁)g/(m₂+m₁)

If m₁=3,1 kg, m₂ = 4,4 kg blir a = 1,7 m/s²

Figure 6-12

Swinging a ball in a circle

6-5 Circular Motion

An object moving in a circle must have a force

acting on it; otherwise it would move in a straight line.

The direction of the force is towards the center of the circle.

Table 6-2

sin θ / θ for values of θ Approaching Zero

6-5 Circular Motion

Some algebra gives us the magnitude of the acceleration, and therefore the force, required to keep an object of mass m moving in a circle of radius r.

The magnitude of the force is given by:

(6-15)

6-5 Circular Motion

This force may be provided by the tension in a

string, the normal force, or friction, among others.

Example 6-8 (p.163) Rounding a Corner (m= 1200 kg, r = 45 m, μ_s = 0,72, v_max = ?)

Example

f_s = ma_x (f_s = μ_sN)

N - mg = 0

μ_sN = mv_max²/r (when μ_s is maximum)

v_max = (0,82•9,81•45)^1/2 = 19 m/s

Photo 6-7 Banking

Example 6-9 Bank on It (p.165)

(m= 900 kg, v = 20,5 m/s, r = 85 m, θ = ?)

Example

N sinθ = ma_x = ma_CP= mv²/r N cosθ - W = 0

Division gives

tanθ = mv²/mgr = v²/gr = (20,5)²/(9,81•85,0) = 0,504 θ = arctan(0,504) = 26,7 º

Active Example 6-4 Find the Normal Force

Active example

N - mg = ma_y a_y = v²/r

N = mg + mv²/r = m(9,81 + 4,47) = 1140 N

Figure 6-14

Simplified top view of a centrifuge in operation

Exercise

(Find the magnitude of the acceleration when v = 89,3 m/s, r = 8,50 cm)

a_cp = v²/r = (89,3)²/(0,0850) = 93800 m/s² = 9560 g

6-5 Circular Motion

An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:

Summary of Chapter 6

• Friction is due to microscopic roughness.

• Kinetic friction:

• Static friction:

• Tension: the force transmitted through a string.

• Force exerted by an ideal spring:

Summary of Chapter 6

• An object is in translational equilibrium if the net force acting on it is zero.

• Connected objects have the same acceleration.

• The force required to move an object of mass m in a circle of radius r is: