Chapter 5 Applying Newton's Laws Flashcards

Applying Newton's Laws

Finding the Net Force

• Free-body diagrams are used to visualize forces acting on an object.
• The net force is the vector sum of all forces acting on the object.

Chapter 5 Preview: Dynamics Problems

• Newton’s laws relate forces to motion, enabling the solution of dynamics problems.
• The acceleration of an object is determined by the forces acting on it.

Equilibrium

Static Equilibrium

• An object at rest is in static equilibrium.

Dynamic Equilibrium

• An object moving in a straight line at a constant speed is in dynamic equilibrium.

Net Force in Equilibrium

• In both types of equilibrium, the net force acting on the object is zero.
• If no net force acts on an object in the xy-plane, there is no net force in either the x- or y-components.

Finding Forces on an Orangutan (Example)

• An orangutan weighing 500 N hangs from a vertical rope. The tension in the rope can be found using equilibrium conditions.
• Free-body diagrams are essential for visualizing forces.

Finding Tension in a Rope Towing a Car (Example)

• A car with a mass of 1500 kg is towed at a steady speed by a rope at an angle from the horizontal.
• A friction force of 320 N opposes the car’s motion.
• The tension in the rope can be calculated considering the equilibrium in both horizontal and vertical directions.

Example Problem: Block on a Rope

• A 100-kg block (weight 980 N) hangs on a rope.
• Find the tension in the rope when the block is stationary and when it’s moving upward at a steady speed of 5 m/s.
• In both cases, the tension equals the weight because the acceleration is zero.

Readying a Wrecking Ball (Example)

• A wrecking ball weighing 2500 N hangs from a cable.
• It’s pulled back to a 20° angle by a second horizontal cable.
• The tension on the horizontal cable can be calculated using trigonometric relationships.

Dynamics and Newton's Second Law (N2)

• Newton’s Second Law is fundamental for dynamics problems.

Towing a Skier (Example)

• A tow rope pulls a skier (weight 800 N) up a 15° slope at constant speed.
• The friction force on the skier is 90 N.
• The magnitudes of the tension and normal forces can be determined by resolving forces along the slope and perpendicular to it.

Dynamics and Newton’s Second Law

• The forces acting on an object determine its acceleration.
• The object’s motion can be found using kinematic equations.
• Newton’s second law can be expressed as: \vec{F}_{net} = m \vec{a}

Putting a Golf Ball (Example)

• A golfer putts a 46 g ball with a speed of 3.0 m/s.
• Friction exerts a 0.020 N retarding force.
• Determine if the putt will reach the hole 10 m away by calculating the deceleration and distance traveled.

Towing a Car with Acceleration (Example)

• A car with a mass of 1500 kg is towed by a rope at an angle to the horizontal.
• A friction force of 320 N opposes the car’s motion.
• What is the tension in the rope if the car goes from rest to 12 m/s in 10 s?

Example Problem: Accelerating Block

• A 100-kg block with a weight of 980 N hangs on a rope.
• Find the tension in the rope if the block is accelerating upwards.

Example Problem: Ball Pulled by a Rope

• A ball weighing 50 N is pulled back by a rope at an angle.
• Find the tension in the pulling rope.

Example Problem: Sled on Ice

• A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
• It crosses a rough patch of snow that exerts a friction force of 12 N.
• How far does it slide on the snow before coming to rest?

Mass and Weight

Mass

• Mass is a quantity that describes an object’s inertia, its resistance to acceleration.

Weight

• Weight is the gravitational force exerted on an object by a planet, varying from planet to planet.
• Weight is different on Jupiter than on Earth, but mass is the same.

Mass and Weight

• Conversion between force units.
• Correspondence between mass and weight, assuming W = mg.

Apparent Weight

• The weight of an object is the force of gravity on that object.
• Your sensation of weight is due to contact forces supporting you.
• Apparent weight is defined in terms of the force you feel: w{app} = F{contact}.

Apparent Weight

• The only forces acting on the man are the upward normal force of the floor and the downward weight force: \vec{F}_{net} = \vec{n} + \vec{w}.
• Thus n = w_{app} and the man feels heavier than normal when accelerating upward.

Finding a Rider’s Apparent Weight in an Elevator (Example)

• Anjay’s mass is 70 kg. He is standing on a scale in an elevator moving at 5.0 m/s.
• As the elevator stops, the scale reads 750 N.
• Determine if the elevator was moving up or down before stopping and how long it took to come to rest.

Weightlessness

• A person in free fall has zero apparent weight.
• “Weightless” does not mean “no weight.” An object that is weightless has no apparent weight.

Example Problem: Student in an Elevator

• A 50-kg student gets in a 1000-kg elevator at rest.
• As the elevator begins to move, she has an apparent weight of 600 N for the first 3 s.
• How far has the elevator moved, and in which direction, at the end of 3 s?

Normal Forces

• Normal forces are always perpendicular to the surface of contact.
• They help to balance other forces and keep objects from penetrating surfaces.

Normal Forces

• An object at rest on a table experiences an upward normal force.
• This force is perpendicular to the surface of contact.
• The normal force adjusts itself so that the object stays on the surface without penetrating it.

Normal Force on a Pressed Book (Example)

• A 1.2 kg book lies on a table and is pressed down from above with a force of 15 N.
• What is the normal force acting on the book from the table below?

Normal Forces on an Incline

• The normal force always points perpendicular to the surface.
• The weight force always points straight down and can be decomposed into x- and y-components.
• When we rotate the x-axis to match the surface, the angle between w and the negative y-axis is the same as the angle of the slope.
• wx = w \sin{\theta}, wy = w \cos{\theta}

Normal Forces on an Incline (Common Mistakes)

• The normal force is always perpendicular to the surface of contact.
• The weight always points straight down.

Acceleration of a Downhill Skier (Example)

• A skier slides down a steep slope.
• Friction is much smaller than the other forces.
• What is the skier’s acceleration?

Friction

• Friction opposes motion between surfaces.

Static Friction

• Static friction is the force that a surface exerts on an object to keep it from slipping.
• The direction of static friction opposes the direction the object would move if there were no friction.

Static Friction

• The box is in static equilibrium.
• The static friction force must exactly balance the pushing force.

Static Friction

• The harder the woman pushes, the harder the friction force from the floor pushes back.
• The static friction force has a maximum possible magnitude: f{s{max}} = \mus n, where \mus is the coefficient of static friction.

Rules for Static Friction

• The direction of static friction opposes motion.
• The magnitude f_s of static friction adjusts itself so that the net force is zero.
• The magnitude of static friction cannot exceed the maximum value f{s{max}} = \mu_s n.
• If the friction force needed to keep the object stationary is greater than f{s{max}}, the object slips and starts to move.

Kinetic Friction

• Kinetic friction has a nearly constant magnitude given by: fk = \muk n, where \mu_k is the coefficient of kinetic friction.
• The magnitude of the kinetic friction force does not depend on how fast the object is sliding.

Rolling Friction

• A wheel rolling on a surface experiences friction, but not kinetic friction.
• The portion of the wheel that contacts the surface is stationary.
• Rolling friction opposes the motion, defined by a coefficient of rolling friction.

Working with Friction Forces

• Static Friction: fs \leq f{s{max}} = \mus n, direction opposes potential motion.
• Kinetic Friction: fk = \muk n, direction opposes motion.
• Rolling Friction: fr = \mur n, direction opposes motion.

Finding the Force to Slide a Sofa (Example)

• Carol wants to move her 32 kg sofa using sofa sliders with \mu_k = 0.08.
• She pushes the sofa at a steady 0.40 m/s.
• How much force does she apply to the sofa?

Causes of Friction

• All surfaces are rough on a microscopic scale.
• When two objects are in contact, high points become jammed against each other.
• The amount of contact depends on how hard the surfaces are pushed together.

Example Problem: Car Stopping

• A car traveling at 20 m/s stops in a distance of 50 m.
• Assume constant deceleration.
• Coefficients of friction between a passenger and the seat are \mus = 0.5 and \muk = 0.3.
• Will a 70-kg passenger slide off the seat if not wearing a seat belt?

Drag

• Drag is a force that opposes motion through a fluid.

Drag

• The drag force is:
  • Opposite in direction to the velocity.
  • Increases in magnitude as the object’s speed increases.
• Drag force is more complex than friction and originates from inertial and viscous properties.
• The Reynolds number helps determine which cause of drag applies.

Reynolds Number

• When a thrown baseball moves through the air, most of the drag occurs due to inertial forces proportional to \rho v^2 L, where \rho is the fluid density, v is the velocity, and L is the object's size.
• When a small sphere falls through honey, drag results from viscous forces proportional to \eta v, where \eta is the viscosity of the fluid.

Reynolds Number

• The Reynolds number is the ratio of inertial force to viscous force: Re = \frac{\rho v L}{\eta}.
• High Reynolds number (over 1000) is dominated by inertial forces.
• Low Reynolds number (less than 1) is dominated by viscosity.
• Table of fluid densities and viscosities for air, ethyl alcohol, olive oil, water, and honey at different temperatures.

Drag at High Reynolds Number

• For high Reynolds numbers, the drag force for motion through a fluid at speed v is: F{drag} = \frac{1}{2} CD \rho A v^2, for Re > 1000.
• A is the cross-section area of the object.
• C_D is the drag coefficient and depends on the object's shape.
• The drag force is proportional to the square of the object’s speed.

Drag at High Reynolds Number (Typical Drag Coefficients)

• Examples of drag coefficients for commercial airliner, Toyota Prius, pitched baseball, racing cyclist, and running person.

Drag at High Reynolds Number

• Cross-section areas and drag coefficients for a sphere and a cylinder illustrate how shape affects drag.

The Drag Coefficient of a Swimming Penguin (Example)

• A gliding 4.8 kg Gentoo penguin has an acceleration of -0.15 m/s^2 when its speed is 1.60 m/s.
• Its frontal area is 0.075 m^2.
• Find the penguin’s drag coefficient.

Terminal Speed

• Just after an object is released from rest, its speed is low and the drag force is small.
• As it falls farther, its speed and hence the drag force increases and eventually the drag force has exactly the same magnitude as gravity.

Terminal Speed

• The steady, unchanging speed at which drag exactly counterbalances an applied force is called the object’s terminal speed.

Terminal Speeds of a Man and a Mouse (Example)

• A 75 kg skydiver and his 0.020 kg pet mouse are falling after jumping from a plane.
• Find the terminal speed of each.

Drag at Low Reynolds Number

• Low Reynolds number objects are mostly small objects with a Re < 1.
• For low Reynolds number, the magnitude of the drag force is proportional to the object’s speed, called linear drag.
• For a spherical object of radius r, the drag force according to Stokes’ law is: F_{drag} = 6 \pi \eta r v, for Re < 1.

Life at Low Reynolds Number

• Examples of objects with low Reynolds numbers include protozoa swimming in pond water and bacteria moving in intercellular water.
• Motion at low Reynolds number is very different.
• After force is applied, terminal speed, v_t, is reached almost instantaneously.
• When force is removed, object also halts almost instantaneously.
• Requires specialized forms of locomotion like propeller-like flagellum of some bacteria.

Measuring the Mass of a Pollen Grain (Example)

• A diameter pollen grain was observed to fall at a rate of 5.3 cm/s.
• What is the mass in nanograms of this grain?

Interacting Objects

• Forces always occur in action/reaction pairs.

Interacting Objects

• Newton’s third law states:
  • Every force occurs as one member of an action/reaction pair of forces.
  • The two members of the pair always act on different objects.
  • The two members of an action/reaction pair point in opposite directions and are equal in magnitude.

Objects in Contact

• To analyze block A’s motion, identify all forces acting on it and draw its free-body diagram.
• Repeat for block B.
• Forces on A and B are not independent: \vec{F}{A on B} = -\vec{F}{B on A} (action/reaction pair, same magnitude).
• Because the two blocks are in contact, their accelerations must be the same: aA = aB.
• We can’t solve for the motion of one block without considering the motion of the other block.

Pushing Two Blocks (Example)

• A 5.0 kg block A is pushed with a 3.0 N force. In front of this block is a 10 kg block B; the two blocks move together.
• What force does block A exert on block B?

Ropes and Pulleys

• Ropes and pulleys redirect forces and can provide mechanical advantage.

Ropes

• The box is pulled by the rope, so the box’s free-body diagram shows a tension force \vec{T}.
• We make the massless string approximation.
• Newton’s second law for the rope is thus T = F.

Ropes

• Generally, the tension in a massless string or rope equals the magnitude of the force pulling on the end of the string or rope.
• A massless string or rope “transmits” a force undiminished from one end to the other.
• If you pull on one end of a rope with force F, the other end of the rope pulls with a force of the same magnitude F.
• The tension in a massless string or rope is the same from one end to the other.

Example Problem: Wooden Box Pulled by a Rope

• A wooden box, with a mass of 22 kg, is pulled at a constant speed with a rope that makes an angle of with the wooden floor.
• What is the tension in the rope?

Pulleys

• The tension in a massless string is unchanged by passing over a massless, frictionless pulley.
• We’ll assume such an ideal pulley for problems in this chapter.

Lifting a Stage Set (Example)

• A 200 kg set is stored in the loft above the stage. The rope holding the set passes up and over a pulley, then is tied backstage.
• A 100 kg stage-hand is told to lower the set. When he unties the rope, the set falls and the stagehand is hoisted into the loft.
• What is the stagehand’s acceleration?