Lecture Outlines Chapter 6 Physics 3 rd Edition

Lecture Outlines Chapter 6 Physics, 3 rd Edition James S. Walker © 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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 kinetic friction. is called the coefficient of

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, μk= 0, 26, t =? ) Making a Big Splash

6 -1 Frictional Forces 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, 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, μs=? ) 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, 22 N, 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, T 1=? , T 2=? ) 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 = m 1 a 1 = m 1 a T = m 2 a 2 = m 2 a (6 -7) F = m 1 a + m 2 a = (m 1+m 2)a a = F/(m 1+m 2) (6 -8) T = m 2 F/(m 1+m 2) (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 6 -4 Tension in the string (Attention, the box slides without friction!) T = m 1 a W 2 - T = m 2 a W 2 = m 2 g W 2 = (m 1+m 2)a a = m 2 g/(m 1+m 2) T = m 1 m 2 g/(m 1+m 2)

Example 6 -7 (p. 161) Atwood’s Machine

Example 6 -7 Atwood’s Machine 6 -4 T - m 1 g = m 1 a m 2 g - T = m 2 a Eliminate T by adding the two equations g(m 2 - m 1) = a(m 2+m 1) a = (m 2 -m 1)g/(m 2+m 1) If m 1=3, 1 kg, m 2 = 4, 4 kg blir a = 1, 7 m/s 2

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, μs = 0, 72, vmax = ? )

Example 6 -8 Rounding a Corner fs = max (fs = μs. N) N - mg = 0 μ s. N = mvmax 2/r vmax = (0, 82 • 9, 81 • 45)1/2 = 19 m/s (when μs is maximum)

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 6 -9 Bank on it N sinθ = max = ma. CP= mv 2/r N cosθ - W = 0 Division gives tanθ = mv 2/mgr = v 2/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 6 -4 Find the Normal Force (v = 17, 0 m/s, m = 80, 0 kg, r = 65, 0 m) N - mg = may ay = v 2/r N = mg + mv 2/r = m(9, 81 + 4, 47) = 1140 N

Figure 6 -14 Simplified top view of a centrifuge in operation

Exercise 6 -1 (p. 167) (Find the magnitude of the acceleration when v = 89, 3 m/s, r = 8, 50 cm) acp = v 2/r = (89, 3)2/(0, 0850) = 93800 m/s 2 = 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: