Chapter Four Holt Physics Forces and the Laws of Motion
Physics Force and the study of dynamics 1.Forces - a. Force - a push or a pull. It can change the motion of an object; start or stop movement; and, change shape of object. -Originally described by Sir Isaac Newton as F=ma -Dynamics= the connection between force and motion
Physics Force and the study of dynamics b. Four basic types (1) gravitational - weakest, attractive force between objects (2) electromagnetic - results from basic property of particles. Large compared to gravitational (3) strong nuclear forces - holds nucleus together, limited in range. (4) weak nuclear - deals with radiation.
Physics Force and the study of dynamics c. Also classified by how they act (1) contact forces act due to physical contact between objects. Push/pull. (2) field forces do not require contact - such as gravitational forces, electromagnetic forces, nuclear forces fields - regions around an object that are influenced by a characteristic of the object - mass, magnetic, etc.
Physics Force and the study of dynamics d. Unit of Force in the SI system is the Newton (N) (1) one Newton is the force required to give a mass of one kilogram an acceleration of one meter per second squared. (2) 1 N = 1 kg-m / s 2 derived unit (3) 1 dyne = 1 g-cm/sec 2 cgs system (4) 1 lb = 1 slug-ft/sec 2 British Engineering System
Physics Force and the study of dynamics (5) Conversions (a) 1 lb = 4.448 N (b) 1 N = 0.225 lb c) 1 N = 10 5 dynes Example: What is the Newton force of a 2 lb bar? 2lbs * 4.448N = 8.896N 1 1lbs
2.Free-Body Diagrams a. A technique to use in solving problems (1) Sketch object under consideration - Rather than drawing a pictorial representation of an object just show it as a box, square, or circle. Physics Force and the study of dynamics
(2) Draw and label all external forces acting on object as vector arrows (a) Assume a direction for each force. If your selection ends up negative(-) means it goes the other way (b) Assume that all forces act at the center of mass of an object. No matter where they act they are shown as acting at the center. (3) forces that the object exerts on other objects, its surroundings, are not shown. Only those that act ON the object. Label arrows/forces The size of the vector arrow indicates the magnitude of the force. Physics Force and the study of dynamics
Overview of Types of Forces - Labels of free-body diagrams 1. Applied Force= force applied to an object by a person or another object. Example: a person pushes a desk across the room, applied force acting upon the desk. 2. Gravity=force by which the earth, moon, or any other planet/massive object attracts another object toward itself. Downward pull towards the center, on earth. 3. Normal Force= support force exerted upon an object that is in contact with another stable object. Example: an object is resting on a surface, then the surface is exerting an upward force upon the object in order to support its weight. 4. Friction Force= force exerted by a surface as an object moves across it. The two types of friction are kinetic and static friction. 5. Air Resistance= acts upon objects as they travel through the air. Often opposes the motion of an object, but is frequently neglected due to negligible magnitude. Example: the force that slows a skydiver while he is falling. 6. Tension=transmitted through a string, rope, cable or wire when pulled tight by forces from opposite ends 7. Spring Force= force exerted by a compressed or stretched spring upon any object that is attached to it. The object that is compressed or stretched is also acted upon by a restoring force that restores it to rest or equilibrium position (Hooke s Law)
Labeling forces - Examples F n F f F app F g F n F n F app F g F g F n F tens F n F g F g
Physics Force and the study of dynamics Use a free-body diagram to determine the net external force acting on the object. The size of the vector arrow indicates the magnitude of the force.
Constructing free-body diagrams - Examples
Free-body diagram - Practice Draw a force diagram of a crash-test dummy in a car at the moment of collision. For this problem, assume that the forces acting on the car are 19 600 N downward, 17 800 N forward, and 25 000 N backward. F net x = 17,800N + (-25,000N) = -7200N Rtor 7200N Lt F net y = 19,600N + (-19,600N) = 7200 Draw a free-body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.
Newton's First Law - Law of Inertia a. Net external force - combination of all forces acting on an object, a vector sum. b. Two parts - body at rest, body in motion c. An object at rest will remain at rest, and an object in uniform motion remains in uniform motion, unless acted upon by a net external force.
d. inertia depends on object's mass, so mass is the measure of inertia - which is the tendency of an object not to accelerate. (1) mass is the amount of matter in an object (2) mass also measures the amount of inertia an object has
Newton's Second Law a. F = m a but more easily understood by a = F / m b. acceleration is directly proportional to the net external force applied to the object and inversely proportional to mass of object. c. second law is a vector equation - direction of acceleration is the same direction as the net force
Newton s 2 nd Law Example
Newton's Third Law - Action/Reaction a. forces always occur in pairs. b. If two bodies interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1, and these forces are opposite in direction. c. Two bodies, two forces - key is to recognize what the forces are
d. When the net external force acting on an object is zero, its acceleration is zero. (1) NOTE: key concept is net external force - if it is zero then the object s motion is not changing. (2) NOTE: motion can be occurring if the object has a zero net force acting - but that motion does not change in magnitude or direction.
g. Equilibrium - net force acting on an object is zero 1. Forces are acting but they cancel each other out 2. Object is at rest or is moving at a constant velocity - which means its speed is not changing nor is its direction of motion.
6. Mass and Weight a. weight - due to gravitational force (1) w = m g (F=ma) (2) direction is downward b. mass - amount of matter in an object. Two types: inertial and gravitational (1) inertial mass - measured using m = F / a (a) force needed to accelerate an object gives mass (b) difficult to do - frictionless surface and measuring acceleration
(2) gravitational mass (a) measured using a pan balance to compare weights of two objects (b) unknown mass on one side, known mass on the other (c) essentially different concepts, always numerically equal. Equality of the two types is called the "equivalence principle" (d) weight is a vector, mass is a scalar
7. Normal Force - F N a. A contact force exerted by one object on another in a direction perpendicular to the surface of contact. b. normal is a mathematical term meaning the force is perpendicular. c. a reaction force - but one that does act on an object.
8. Friction a. A force that opposes the motion of two objects that are touching each other b. An electromagnetic force resulting from temporary attractions between the contact points of the two surfaces c. friction always acts parallel to the surfaces in contact and in a direction opposing motion
d. Two main types (1) Static Friction - force of friction resisting the start of motion. Varies in magnitude depending on the applied force trying to start motion. (2) Kinetic (Sliding) Friction - force resisting existing motion (3) Static friction is always larger than sliding friction (4) We will work mainly with kinetic friction
e. Coefficient of Friction (µ) = F f / F N (1) F N represents the normal force. The force pulling the surfaces together. It is always perpendicular to the surfaces in contact. (2) F f is the force of sliding friction. It is parallel to the surfaces in contact and in the opposite direction from motion
f. Example: a box being pulled at a constant speed over a level surface. F N F f F Applied F weight
(1) since constant speed - no acceleration, forces are balanced (2) not moving off surface so sum of forces in y direction is equal to 0 (3) force of friction depends only on the nature of the surfaces in contact and Normal Force. (4) coefficient of friction (µ) is independent of the surface areas in contact and the velocity of the object
g. Example: A smooth wooden block is placed on a smooth wooden tabletop. A force of 14 N is necessary to keep the 40N object moving at a constant speed. (1) Find coefficient of friction (µ) (2) If a 20.0N weight is placed on the block what force will be required to keep the block and weight moving at a constant velocity.
Given: W = 40N F A = 14N F f = 14N since constant velocity F N = 40N, since object not being raised F f = µ F N F =14N F N = 40N F f =14N W = 40N
9. Net Force Causes Acceleration a. Net force is the vector sum of all forces acting on a body F net = F applied + F friction b. Ex: Consider a mass of 50 kg sitting on a frictionless surface. A force of 100N is applies. Find a.
Given: m =50kg F APPLIED = 100N F friction = 0 F = m a A = F / m a = 2 m/s 2 c. If in the above problem µ is 0.2, find a.
F friction = µ F N = 0.2 (50kg x 9.8 m/s 2 ) and it opposes motion of applied force F Net = 100N + (-98N) =2N a=.04m/s 2 F applied = 100N F friction = 98N
d. Student Problem. A shopper in a supermarket cart pushes a loaded cart with a horizontal force of 10.0N. If the cart has a mass of 30.0 kg, how far will it move in 3 sec, starting from rest if (1) you ignore friction (2) if the shopper places his 30.0 N child in the cart before he pushes it?
Answer: F = m a a = 1/3 m/ s 2 s = v 0 t + 1/2 at 2
e. Elevator Problems (1) elevator at rest - weight reading on the scale is from the normal force of the scale pushing back up on the object which is pushing down due to gravity (2) with elevator at rest, a Y = 0 and ΣF Y = 0 scale weight normal force ΣF Y = 0 = F N - W F N = W
(3) elevator moving up, a Y = +n & ΣF Y =m a Y weight ΣF Y = m a Y = F N - W F N = apparent weight= = W + m a Y normal force
(4) elevator moving down, a Y = -n & ΣF Y =-m a Y weight ΣF Y = -m a Y = F N - W F N = apparent weight= = W - m a Y normal force
10. Terminal Velocity a. constant velocity when F drag = W b. Air resistance is a frictional force c. depends on the density of air, size and shape of the object, and speed of motion
11. Force at an angle problems Derek leaves his physics book on top of a drafting table that is inclined at a 35 angle. The free- body diagram in Figure 4-8 shows the forces acting on the book. Find the net external force acting on the book, and determine whether the book will remain at rest in this position.
1. Define the problem and identify the variables. Given: F gravity on book = 22 N F friction = 11 N F table on book = 18 N Unknown: F net =? 2. Select a coordinate system, and apply it to the free-body diagram. Choose the x-axis parallel to and the y- axis perpendicular to the incline of the table, as shown in Figure 4-9. This is the most convenient coordinate system because only the force of gravity on the book needs to be resolved into x and y components. All other forces are either along the x- axis or the y-axis.
3. Find the x and y components of all vectors. Draw a sketch as shown in Figure 4-10 to help find the x component and y component of the vector F gravity-on-book Gravity acts at a 90 angle to the surface of Earth, and the x- axis is at a 35 angle to Earth s surface. Therefore, θ, the angle between the gravity vector and the x- axis, is the third angle in this right triangle and is equal to 55.
4. Find the net external force in both the x and y directions. Figure 4-12 shows another freebody diagram of the book, now with forces acting only along the x- and y- axes.
5. Find the net external force. The net force in the y direction is equal to zero, so the net external force is equal to the net force in the positive x direction, 2 N positive. 6. Evaluate your answer. The net external force acts on the book in the downhill direction. Therefore, the book will experience an acceleration in the downhill direction, and it will slide off the table, as shown Figure 4-13.
12. Two-body Problems a. To solve a problem involving two or more bodies write F = ma for each body separately, having first decided which direction of motion you want to designate as positive. b. Both objects will have same acceleration, just different directions
Equilibrium Problems c. Example: Two masses are tied to opposite ends of a massless rope, and the rope is hung over a massless and frictionless pulley. Find the acceleration of the masses.
(1) 15 kg mass falls, turns CW, so designate down as positive motion for 15 kg object and upward as positive for 10 kg object. (2) Draw freebody diagrams (3) Note tension is the same throughout the rope 10 kg 15 kg
(4) T - 98N = 10kg (a) 147N - T = 15kg (a) T T two equations with two unknowns T and a w = 98 N W = 147 N