Physics

Chapter 6 Applications of Newton’s Laws

Imron Rosyadi

Introduction

• Newton’s Laws are fundamental to understanding motion.
• This chapter applies these laws to more complex scenarios.
• We’ll explore forces like friction, centripetal force, and drag.

Welcome to Chapter 6! This section is all about applying Newton’s laws to real-world situations, building on what you learned in the previous chapter. We’ll move beyond simple, ideal cases to look at more complex systems involving various types of forces. The goal is to develop a robust problem-solving toolkit.

6.1 Solving Problems with Newton’s Laws

Learning Objectives

By the end of this section, you will be able to:

• Apply problem-solving techniques to complex systems of forces.
• Use kinematics to solve problems involving Newton’s laws.
• Solve more complex equilibrium problems.
• Solve more complex acceleration problems.
• Apply calculus to advanced dynamics problems.

These learning objectives outline what you should be able to do after studying this section. We’ll be integrating concepts from previous chapters, particularly kinematics, to tackle more challenging dynamics problems.

Problem-Solving Strategies

• Identify Principles: List givens, unknowns, and physical principles.
• Sketch Situation: Draw a clear diagram with force arrows.
• Define System: Isolate the system of interest for a free-body diagram.
• Apply Newton’s Second Law: Solve using \(\sum \vec{F} = m\vec{a}\).
• Check Solution: Ensure the answer is reasonable and units are correct.

This is a reiteration of the problem-solving strategy from earlier chapters, but with an emphasis on its application to Newton’s laws. The free-body diagram is crucial here; make sure you understand how to draw one correctly, including only external forces on your system of interest. Remember, if acceleration is zero, you’re dealing with equilibrium. If not, then Newton’s second law, \(\sum \vec{F} = m\vec{a}\), applies directly.

Free-Body Diagrams in Action

Figure 6.2: Lifting a grand piano.

• (a) Sketch: Visualize the scenario.
• (b) Identify Forces: Draw all forces acting on the piano.
• (c) System of Interest: The piano itself.
  • Forces on the piano: Tension (\(\vec{T}\)), Weight (\(\vec{w}\)).
  • Force the piano exerts on the rope (\(\vec{F}_T\)) is not on the system.
• (d) Apply Newton’s 2nd Law: Sum forces to find unknowns.
  • For stationary piano: \(\vec{T} = -\vec{w}\).

This visual example walks through the steps. Notice the distinction between forces acting on the system (piano) and forces the system exerts on the outside world. Only forces acting on the system belong in the free-body diagram. This is a common point of confusion.

Component Equations

• For 2D problems, break forces into components along chosen axes.
• Convenient axes: one parallel to motion (if known), one perpendicular.

\[ \sum F_x = ma_x \]

\[ \sum F_y = ma_y \]

Note

Always check your solution for reasonableness and correct units!

Once you have your free-body diagram, the next step is to apply Newton’s second law by resolving forces into components. Choosing your coordinate system wisely can significantly simplify the problem. For example, on an incline, aligning one axis with the incline is usually best. If the object is accelerating along one direction, say ‘x’, then \(a_y\) would be zero.

Particle Equilibrium: Example 6.1

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure 6.3. Find the tension in each wire, neglecting the masses of the wires.

Figure 6.3: Traffic light suspended by two wires.

• System: Traffic light (mass \(m = 15.0 \text{ kg}\)).
• Forces: \(\vec{T}_1\), \(\vec{T}_2\), \(\vec{w}\) (weight).
• Equilibrium: \(\vec{a} = 0\), so \(\sum \vec{F} = 0\).
  • \(\sum F_x = T_{2x} + T_{1x} = 0 \implies T_1 \cos 30^\circ = T_2 \cos 45^\circ\).
  • \(\sum F_y = T_{1y} + T_{2y} - w = 0 \implies T_1 \sin 30^\circ + T_2 \sin 45^\circ = mg\).
• Solve: Substitute and solve the system of equations.
  • \(T_1 = 109 \text{ N}\)
  • \(T_2 = 133 \text{ N}\)

Here, we’re dealing with static equilibrium, meaning the net force is zero. The key is to break down the tension forces into x and y components. Since the traffic light is not moving (acceleration is zero), both the sum of forces in the x-direction and the sum of forces in the y-direction must be zero. This gives us two equations and two unknowns (\(T_1\) and \(T_2\)), which we can solve simultaneously. Notice that \(T_2\) is greater than \(T_1\) because it’s at a steeper angle, meaning it contributes more vertically and less horizontally, so it needs to be larger to balance the horizontal component from \(T_1\) and still contribute significantly to the vertical balance.

Particle Acceleration: Example 6.2

Drag Force on a Barge

Two tugboats push on a barge at different angles (Figure 6.4). The first tugboat exerts a force of \(2.7×10^5\) N in the x-direction, and the second tugboat exerts a force of \(3.6×10^5\) N in the y-direction. The mass of the barge is \(5.0×10^6\) kg and its acceleration is observed to be \(7.5×10^{−2}\) \(m/s^2\) in the direction shown. What is the drag force of the water on the barge resisting the motion? (Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

Figure 6.4: Tugboats pushing a barge.

• System: Barge (mass \(m = 5.0 \times 10^6 \text{ kg}\)).
• Forces: \(\vec{F}_1 = 2.7 \times 10^5 \text{ N}\) (x-dir), \(\vec{F}_2 = 3.6 \times 10^5 \text{ N}\) (y-dir), \(\vec{F}_D\) (drag).
• Acceleration: \(\vec{a} = 7.5 \times 10^{-2} \text{ m/s}^2\) at \(53.1^\circ\) from x-axis.
• Strategy: Find \(\vec{F}_{app} = \vec{F}_1 + \vec{F}_2\), then apply Newton’s 2nd Law.
  • Magnitude of \(\vec{F}_{app} = \sqrt{(2.7 \times 10^5)^2 + (3.6 \times 10^5)^2} = 4.5 \times 10^5 \text{ N}\).
  • Direction of \(\vec{F}_{app} = \tan^{-1}(3.6/2.7) = 53.1^\circ\).
  • Since drag opposes motion, it acts opposite to \(\vec{F}_{app}\).
• Solve: \(F_{net} = F_{app} - F_D = ma\).
  • \(4.5 \times 10^5 \text{ N} - F_D = (5.0 \times 10^6 \text{ kg})(7.5 \times 10^{-2} \text{ m/s}^2)\).
  • \(F_D = 7.5 \times 10^4 \text{ N}\).

This example introduces particle acceleration, where the net force is non-zero. The key here is to first combine the applied forces from the tugboats into a single resultant force vector. This resultant force gives the direction of the acceleration. Since the drag force opposes motion, it will act in the opposite direction to the resultant applied force. Then, we can apply Newton’s second law in the direction of motion to solve for the unknown drag force.

Example 6.3: Bathroom Scale in an Elevator

Figure 6.5 shows a 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 \(m/s^2\), and (b) if the elevator moves upward at a constant speed of 1 m/s.

Figure 6.5: Man on a scale in an elevator.

• System: Man (mass \(m = 75.0 \text{ kg}\)).
• Forces: Normal force from scale (\(\vec{F}_s\)) upward, Weight (\(\vec{w}\)) downward.
• Newton’s 2nd Law: \(\sum F_y = F_s - w = ma\).
  • \(F_s = mg + ma = m(g+a)\).
• (a) Accelerating Upward: \(a = 1.20 \text{ m/s}^2\).
  • \(F_s = 75.0 \text{ kg} (9.80 \text{ m/s}^2 + 1.20 \text{ m/s}^2) = 825 \text{ N}\).
  • Scale reads more than weight (\(735 \text{ N}\)).
• (b) Constant Upward Speed: \(a = 0 \text{ m/s}^2\).
  • \(F_s = 75.0 \text{ kg} (9.80 \text{ m/s}^2 + 0) = 735 \text{ N}\).
  • Scale reads actual weight.

Tip

What if the elevator accelerates downward at \(1.20 \text{ m/s}^2\)? \(F_s = m(g-a) = 75.0 \text{ kg} (9.80 \text{ m/s}^2 - 1.20 \text{ m/s}^2) = 645 \text{ N}\). Scale reads less than weight.

This is a classic problem that helps differentiate between actual weight and apparent weight. The scale reading measures the normal force exerted on the person. When the elevator accelerates, the normal force changes. When accelerating upwards, the scale must push harder, so the reading is greater. When accelerating downwards, the scale doesn’t need to push as hard, so the reading is less. At constant velocity (or rest), the acceleration is zero, and the scale reads the true weight.

Example 6.4: Two Attached Blocks

Figure 6.6 shows a block of mass \(m_1\) on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass \(m_2\). Find the acceleration of the blocks and the tension in the string in terms of \(m_1\), \(m_2\) ,and g.

Figure 6.6: Block on table, connected to hanging block.

• System: Two blocks, \(m_1\) (on table) and \(m_2\) (hanging).
• Forces on \(m_1\): \(\vec{T}\) (right), \(\vec{w}_1\) (down), \(\vec{N}\) (up).
• Forces on \(m_2\): \(\vec{T}\) (up), \(\vec{w}_2\) (down).
• Constraint: \(a_{1x} = -a_{2y} = a\).
• Newton’s 2nd Law:
  • For \(m_1\): \(\sum F_x = T = m_1 a_x = m_1 a\).
  • For \(m_2\): \(\sum F_y = T - m_2 g = m_2 a_y = -m_2 a\).
• Solve for \(a\) and \(T\):
  • \(a = \frac{m_2 g}{m_1 + m_2}\)
  • \(T = \frac{m_1 m_2}{m_1 + m_2} g\)

Note

Notice that \(T < m_2 g\). If \(T = m_2 g\), the hanging block wouldn’t accelerate.

This problem is an example of a multi-body system where the motion of one object is linked to the other. Drawing separate free-body diagrams for each block is essential. The tension in the string is the same for both blocks (assuming a massless, frictionless string and pulley). The crucial insight is the kinematic constraint: if \(m_1\) moves right by a certain distance, \(m_2\) moves down by the same distance, so their accelerations have the same magnitude. Solving the two equations simultaneously gives us the acceleration and tension in terms of the masses and gravity.

Example 6.5: Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine, which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure 6.7, \(m_1=2.00\) kg and \(m_2=4.00\) kg. Consider the pulley to be frictionless. (a) If \(m_2\) is released, what will its acceleration be? (b) What is the tension in the string?

Figure 6.7: Atwood machine with \(m_1 = 2.00 \text{ kg}\), \(m_2 = 4.00 \text{ kg}\)

• System: Two blocks, \(m_1\) and \(m_2\), connected by a string over a pulley.
• Forces on \(m_1\): \(\vec{T}\) (up), \(\vec{w}_1\) (down).
• Forces on \(m_2\): \(\vec{T}\) (up), \(\vec{w}_2\) (down).
• Constraint: If \(m_2\) accelerates down (\(a\)), \(m_1\) accelerates up (\(a\)).
• Newton’s 2nd Law:
  • For \(m_1\): \(T - m_1 g = m_1 a\).
  • For \(m_2\): \(T - m_2 g = -m_2 a\).
• Solve for \(a\) and \(T\):
  • \(a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(4-2)\text{kg} \cdot 9.8\text{m/s}^2}{(4+2)\text{kg}} = 3.27 \text{ m/s}^2\).
  • \(T = m_1(g+a) = 2\text{kg} (9.8\text{m/s}^2 + 3.27\text{m/s}^2) = 26.1 \text{ N}\).

The Atwood machine is a classic physics setup. Similar to the previous example, we apply Newton’s second law to each mass separately, noting that the tension in the string is the same and the magnitudes of their accelerations are equal but in opposite directions. This allows us to set up a system of two equations with two unknowns (\(T\) and \(a\)). The solution for acceleration is particularly insightful as it shows the unbalanced force divided by the total mass of the system.

Newton’s Laws and Kinematics

• Forces cause accelerations (Newton’s 2nd Law).
• Accelerations change velocities and positions (Kinematics).
• Integrate concepts from both to solve complex problems.

Note

Always list givens, unknowns, and identify involved principles.

Newton’s Laws and Kinematics

Example 6.6: Soccer Player

What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s.

1. What is her average acceleration?

2. What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

• Given: \(v_0 = 0\), \(v_f = 8.00 \text{ m/s}\), \(\Delta t = 2.50 \text{ s}\), \(m = 70.0 \text{ kg}\).
• (a) Average acceleration: \(a = \frac{\Delta v}{\Delta t} = \frac{8.00 \text{ m/s}}{2.50 \text{ s}} = 3.20 \text{ m/s}^2\).
• (b) Average force: \(F_{net} = ma = (70.0 \text{ kg})(3.20 \text{ m/s}^2) = 224 \text{ N}\).

This example shows how kinematics and Newton’s laws are intertwined. Part (a) is purely kinematics: finding acceleration from initial velocity, final velocity, and time. Part (b) then uses that calculated acceleration with Newton’s second law to find the force. This demonstrates a common pattern in physics problems where you might need to use equations from multiple chapters.

Example 6.8: Baggage Tractor

Figure 6.8(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by \(F=(820.0t)\) N, find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Figure 6.8: Baggage tractor and carts.

• System: Tractor (\(m_T = 650.0 \text{ kg}\)), Cart A (\(m_A = 250.0 \text{ kg}\)), Cart B (\(m_B = 150.0 \text{ kg}\)).
• Driving Force: \(F = (820.0t) \text{ N}\).
• Total Mass: \(m_{system} = 650 + 250 + 150 = 1050 \text{ kg}\).
• (a) Speed after 3.00 s:
  • \(\sum F_x = m_{system} a \implies 820.0t = 1050a \implies a = 0.7809t\).
  • Since \(a = dv/dt\), integrate: \(\int_0^v dv = \int_0^{3.00} 0.7809t dt\).
  • \(v = 0.3905t^2 \Big|_0^{3.00} = 0.3905 (3.00)^2 = 3.51 \text{ m/s}\).
• (b) Horizontal force on cable between tractor and cart A:
  • Consider tractor as system of interest (FBD).
  • Forces on tractor: Driving force \(F\) (left), Tension \(T\) from cart A (right).
  • \(\sum F_x = F - T = m_T a\).
  • \(820.0(3.00) - T = (650.0 \text{ kg}) (0.7809 \cdot 3.00)\).
  • \(T = 2460 - 1522 = 938 \text{ N}\).

This problem introduces time-varying forces, requiring calculus. In part (a), we first find the acceleration of the entire system as a function of time, and then integrate that acceleration to find the velocity. For part (b), we switch our system of interest to just the tractor. This allows us to find the tension in the cable connecting the tractor to the first cart. It’s important to remember that the acceleration is the same for all parts of the connected system.

6.2 Friction

• Friction: A force that opposes relative motion between systems in contact.
  • Parallel to contact surfaces, opposes motion/attempted motion.
• Kinetic Friction (\(f_k\)): When objects are moving relative to each other.
• Static Friction (\(f_s\)): When objects are stationary relative to each other.
  • Usually greater than kinetic friction.
  • Responds to applied force up to a maximum limit.

Note

Friction arises from roughness of surfaces and adhesive forces between molecules.

Friction is an incredibly common and important force in our daily lives, often overlooked until we try to move something heavy or walk on ice. It’s not a fundamental force in the same way gravity or electromagnetism is, but rather an emergent property of surfaces in contact. It always acts to oppose the relative motion or tendency of motion between surfaces. Understanding the difference between static and kinetic friction is key.

Magnitudes of Friction

• Static Friction: \(f_s \le \mu_s N\).
  • \(\mu_s\): coefficient of static friction.
  • \(N\): magnitude of normal force.
  • \(f_s\) adjusts to match the applied force until it reaches its maximum \(f_{s(max)} = \mu_s N\).
• Kinetic Friction: \(f_k = \mu_k N\).
  • \(\mu_k\): coefficient of kinetic friction.
  • Always less than or equal to \(f_{s(max)}\) (so \(\mu_k \le \mu_s\)).

Magnitudes of Friction

Figure 6.11: Friction vs. Applied Force.

Typical Coefficients of Friction

System	\(\mu_s\)	\(\mu_k\)
Rubber on dry concrete	1.0	0.7
Wood on wood	0.5	0.3
Steel on steel (dry)	0.6	0.3
Teflon on steel	0.04	0.04

These are the empirical formulas for friction. Remember, static friction is a responsive force. It will only be as large as needed to prevent motion, up to its maximum value. Once that maximum is overcome, the object starts moving, and the friction becomes kinetic, which is usually a constant value determined by \(\mu_k N\). The coefficients of friction (\(\mu_s\) and \(\mu_k\)) are dimensionless quantities that depend on the nature of the surfaces in contact.

Example 6.10: Static and Kinetic Friction

A 20.0-kg crate is at rest on a floor as shown in Figure 6.13. The coefficient of static friction between the crate and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal force \(\vec{P}\) is applied to the crate. Find the force of friction if
(a) \(\vec{P}=20.0\,\text{N}\,\hat{i}\),
(b) \(\vec{P}=30.0\,\text{N}\,\hat{i}\),
(c) \(\vec{P}=120.0\,\text{N}\,\hat{i}\),
and (d) \(\vec{P}=180.0\,\text{N}\,\hat{i}\).

Figure 6.13: Crate on floor, pushed by \(\vec{P}\)

• Given: Crate \(m = 20.0 \text{ kg}\), \(\mu_s = 0.700\), \(\mu_k = 0.600\).
• Normal Force: \(N = w = mg = (20.0 \text{ kg})(9.80 \text{ m/s}^2) = 196 \text{ N}\).
• Max Static Friction: \(f_{s(max)} = \mu_s N = (0.700)(196 \text{ N}) = 137 \text{ N}\).
• Kinetic Friction: \(f_k = \mu_k N = (0.600)(196 \text{ N}) = 118 \text{ N}\).
• Applied Force \(\vec{P}\):
  • (a) \(P = 20.0 \text{ N}\): \(P < f_{s(max)}\), so \(f_s = P = 20.0 \text{ N}\). Crate is at rest (\(a=0\)).
  • (b) \(P = 30.0 \text{ N}\): \(P < f_{s(max)}\), so \(f_s = P = 30.0 \text{ N}\). Crate is at rest (\(a=0\)).
  • (c) \(P = 120.0 \text{ N}\): \(P < f_{s(max)}\), so \(f_s = P = 120.0 \text{ N}\). Crate is at rest (\(a=0\)).
  • (d) \(P = 180.0 \text{ N}\): \(P > f_{s(max)}\), so crate moves. Friction is kinetic.
    • \(f_k = 118 \text{ N}\).
    • \(a_x = \frac{P - f_k}{m} = \frac{180.0 \text{ N} - 118 \text{ N}}{20.0 \text{ kg}} = 3.10 \text{ m/s}^2\).

This example clearly illustrates the behavior of static vs. kinetic friction. Static friction matches the applied force up to its maximum value, keeping the object at rest. Once the applied force exceeds this maximum, the object starts moving, and the friction immediately drops to its kinetic value. This change in the nature of friction is crucial for solving problems where objects transition from rest to motion.

Friction and the Inclined Plane

• Objects on a slope often involve friction.
• Component of weight parallel to slope: \(mg \sin \theta\).
• Component of weight perpendicular to slope (normal force): \(N = mg \cos \theta\).

Friction and the Inclined Plane

Example 6.11: Downhill Skier

A skier with a mass of 62 kg is sliding down a snowy slope at a constant acceleration. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.

Figure 6.14: Skier on a slope.

• Given: Skier \(m = 62 \text{ kg}\), slope angle \(25^\circ\), \(f_k = 45.0 \text{ N}\).
• Normal Force: \(N = mg \cos 25^\circ = (62 \text{ kg})(9.80 \text{ m/s}^2)(\cos 25^\circ) = 551 \text{ N}\).
• Coefficient of Kinetic Friction:
  • \(f_k = \mu_k N \implies \mu_k = \frac{f_k}{N}\).
  • \(\mu_k = \frac{45.0 \text{ N}}{551 \text{ N}} = 0.082\).
• Constant Velocity Condition: If an object slides down an incline at constant velocity, then \(mg \sin \theta = f_k = \mu_k N = \mu_k mg \cos \theta\).
  • This simplifies to \(\mu_k = \tan \theta\).

When dealing with inclined planes, choosing a coordinate system aligned with the plane (one axis parallel, one perpendicular) greatly simplifies the problem. The weight vector needs to be resolved into components along these new axes. The normal force is then equal to the perpendicular component of the weight. This example shows how to find the coefficient of kinetic friction if the friction force is known, and then derives a useful shortcut formula for when an object slides at constant velocity.

6.3 Centripetal Force

• An object in circular motion is always accelerating (changing direction).
• This acceleration is centripetal acceleration (\(\vec{a}_c\)), directed towards the center of the circle.
• Magnitude: \(a_c = \frac{v^2}{r} = r\omega^2\).
• Centripetal Force (\(\vec{F}_c\)): The net force causing centripetal acceleration.
  • Magnitude: \(F_c = m a_c = \frac{m v^2}{r} = m r \omega^2\).
  • Direction: Always towards the center of curvature, perpendicular to velocity.

Important

Centripetal force is not a new type of force; it is the net force resulting from other forces (e.g., tension, gravity, friction) that causes circular motion.

Recall from kinematics that circular motion implies acceleration, even if speed is constant. This acceleration is always directed towards the center, hence “centripetal.” According to Newton’s second law, any acceleration must be caused by a net force. This net force is what we call centripetal force. It’s crucial to understand that centripetal force isn’t a fundamental force like gravity; it’s the role that another force or combination of forces plays in causing circular motion.

Example 6.15: Cars on a Flat Curve

1. Calculate the centripetal force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 m/s.

2. Assuming an unbanked curve, find the minimum static coefficient of friction between the tires and the road, static friction being the reason that keeps the car from slipping (Figure 6.21).

Figure 6.21: Car on an unbanked curve.

• Given: Car \(m = 900.0 \text{ kg}\), radius \(r = 500.0 \text{ m}\), speed \(v = 25.00 \text{ m/s}\).
• (a) Centripetal force:
  • \(F_c = \frac{m v^2}{r} = \frac{(900.0 \text{ kg})(25.00 \text{ m/s})^2}{500.0 \text{ m}} = 1125 \text{ N}\).
• (b) Minimum static coefficient of friction:
  • On an unbanked curve, static friction provides the centripetal force.
  • \(F_c = f_s = \mu_s N\).
  • Normal force \(N = mg\) (on level ground).
  • So, \(\frac{m v^2}{r} = \mu_s mg\).
  • \(\mu_s = \frac{v^2}{rg} = \frac{(25.00 \text{ m/s})^2}{(500.0 \text{ m})(9.80 \text{ m/s}^2)} = 0.13\).

This example shows friction acting as the centripetal force. For a car to turn on a flat road, there must be a force pulling it towards the center of the turn. This force is static friction between the tires and the road. If the required centripetal force exceeds the maximum static friction, the car will skid. Notice that the mass of the car cancels out when solving for the coefficient of friction, meaning the minimum coefficient required is independent of the car’s mass for a given speed and radius.

Banked Curves

• Roads are often banked (sloped) to help cars negotiate curves.
• Ideally Banked Curve: An angle \(\theta\) where friction is not needed.
  • Horizontal component of normal force provides \(F_c\).
  • Vertical component of normal force balances weight.

Banked Curves

Figure 6.22: Car on a banked curve.

• Forces: Weight (\(\vec{w}\)) down, Normal force (\(\vec{N}\)) perpendicular to road.
• Component Equations:
  • \(\sum F_x = N \sin \theta = \frac{m v^2}{r}\)
  • \(\sum F_y = N \cos \theta - mg = 0 \implies N \cos \theta = mg\)
• Ideal Banking Angle (\(\theta\)):
  • Divide the two equations: \(\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg}\).
  • \(\tan \theta = \frac{v^2}{rg}\).
  • \(\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)\).

Banked curves allow for safer turns, especially at higher speeds. The banking provides a horizontal component of the normal force, which acts as the centripetal force. In an ideally banked curve, this horizontal component is exactly what’s needed for a specific speed, meaning no friction is required. This formula for the ideal banking angle shows that steeper banks are needed for higher speeds and sharper turns (smaller radius).

6.4 Drag Force and Terminal Speed

• Drag Force (\(F_D\)): Resistance force on an object moving through a fluid (gas or liquid).
  • Always opposes motion.
  • Magnitude depends on shape, size, velocity, and fluid properties.
• For large objects at moderate/high speeds: \(F_D = \frac{1}{2} C \rho A v^2\).
  • \(C\): drag coefficient (dimensionless, empirically determined).
  • \(\rho\): density of the fluid.
  • \(A\): cross-sectional area of the object facing the fluid.
  • \(v\): speed of the object.
• Terminal Velocity (\(v_T\)): Constant velocity reached when drag force equals gravitational force (net force = 0).

Note

Aerodynamic shaping is used to reduce drag in vehicles and sports equipment.

Drag force is another resistive force, but unlike simple friction, its magnitude depends on velocity. For many practical applications, it’s proportional to \(v^2\). This force is crucial in understanding things like skydiving, falling rain, and even the design of cars and airplanes. When an object falls, gravity pulls it down, but drag pushes up. As speed increases, drag increases, until it balances gravity. At this point, the object stops accelerating and reaches its terminal velocity.

Terminal Velocity Calculation

• At terminal velocity, the net force is zero: \(\sum F_y = mg - F_D = 0\).
• So, \(mg = F_D = \frac{1}{2} C \rho A v_T^2\).
• Solving for terminal velocity: \(v_T = \sqrt{\frac{2mg}{C \rho A}}\).

Terminal Velocity Calculation

Example 6.17: Terminal Velocity of a Skydiver

Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position. Assume that in the spread-eagle position, the diver has a cross-sectional area of 0.70 \(m^2\) .

• Given: Skydiver \(m = 85 \text{ kg}\), spread-eagle position \(A = 0.70 \text{ m}^2\), \(C = 1.0\) (from table), \(\rho_{air} = 1.21 \text{ kg/m}^3\).
• \(v_T = \sqrt{\frac{2(85 \text{ kg})(9.80 \text{ m/s}^2)}{(1.0)(1.21 \text{ kg/m}^3)(0.70 \text{ m}^2)}}\).
• \(v_T = 44 \text{ m/s}\).

Tip

Heavier objects have higher terminal velocities because they need a greater drag force to balance their weight, which requires higher speeds.

This formula for terminal velocity is a direct result of balancing the gravitational force with the drag force. It shows that heavier objects with smaller cross-sectional areas and lower drag coefficients will have higher terminal velocities. This explains why a mouse can survive a long fall, but a human cannot (as beautifully described by J.B.S. Haldane).

Calculus of Velocity-Dependent Frictional Forces

• When frictional force depends on velocity (e.g., \(f_R = -bv\)), Newton’s 2nd Law becomes a differential equation.
• For an object falling through a liquid with \(f_R = -bv\):
  • \(mg - bv = m \frac{dv}{dt}\).
  • Integrating yields velocity as a function of time: \(v(t) = \frac{mg}{b}(1 - e^{-bt/m})\).
• Terminal velocity: As \(t \to \infty\), \(v(t) \to \frac{mg}{b} = v_T\).
• Position as a function of time: \(y(t) = \frac{mg}{b}t + \frac{m^2g}{b^2}(e^{-bt/m} - 1)\).

For situations where the resistive force is linear with velocity (like Stokes’ Law for small objects at low speeds), the problem requires solving a differential equation. This slide provides the derived solutions for velocity and position as functions of time. It’s important to recognize when these calculus-based approaches are necessary, particularly when the force is not constant. Note how the terminal velocity naturally emerges from these solutions as time approaches infinity.

Key Takeaways

• Problem-Solving: Systematic approach involving FBDs, Newton’s Laws, and checking reasonableness.
• Equilibrium vs. Acceleration: Distinguish between zero net force and non-zero net force scenarios.
• Friction: Static (\(f_s \le \mu_s N\)) prevents motion, kinetic (\(f_k = \mu_k N\)) opposes motion.
• Centripetal Force: Any force or combination of forces causing circular motion (\(F_c = mv^2/r\)).
• Drag Force: Air/fluid resistance (\(F_D = \frac{1}{2} C \rho A v^2\)), leads to terminal velocity.
• Calculus Applications: Necessary for forces that vary with time or velocity.

This summarizes the most important concepts covered in Chapter 6. Remember to apply the problem-solving strategy to every problem. Understand the nature of each force we discussed, especially how they manifest as centripetal force, and when to use the constant friction formulas versus the velocity-dependent drag force equations.

Key Equations

Equation	Description
\(\sum F = ma\)	Newton’s Second Law
\(f_s \le \mu_s N\)	Magnitude of Static Friction
\(f_k = \mu_k N\)	Magnitude of Kinetic Friction
\(F_c = \frac{mv^2}{r}\)	Centripetal Force (linear velocity)
\(F_c = mr\omega^2\)	Centripetal Force (angular velocity)
\(\tan \theta = \frac{v^2}{rg}\)	Ideal Banking Angle
\(F_D = \frac{1}{2} C \rho A v^2\)	Drag Force (high speeds)
\(v_T = \sqrt{\frac{2mg}{C \rho A}}\)	Terminal Velocity (high speeds)
\(f_R = -bv\)	Drag Force (low speeds / Stokes’ Law)
\(v(t) = \frac{mg}{b}(1 - e^{-bt/m})\)	Velocity with linear drag (calculus)

This table provides a concise summary of the key mathematical relationships covered in the chapter. Familiarize yourself with each equation and the conditions under which it applies.

Key Terms

Term	Definition
Centripetal Force	Any net force causing uniform circular motion. Always directed towards the center of the circular path.
Coriolis Force	Fictitious force arising in a rotating frame of reference, causing objects to appear to deflect perpendicular to their motion.
Drag Force	A force that opposes the motion of an object through a fluid. Its magnitude depends on velocity, shape, and fluid properties.
Friction	A force that opposes relative motion or attempted motion between systems in contact.
Kinetic Friction	Friction that acts between systems in contact when they are moving relative to one another.
Normal Force	The perpendicular contact force that a surface exerts on an object that is pressing on it.
Static Friction	Friction that acts between systems in contact when they are stationary relative to one another.
Terminal Velocity	The constant velocity achieved by a falling object when the drag force equals the force of gravity, resulting in zero net force.

This glossary defines the important terms introduced in this chapter. A clear understanding of these terms is essential for mastering the concepts.