Forces In One Dimension Physics: Chapter 4

4-1: Force In physics, a force is a push or a pull. Unbalanced forces  Acceleration Acceleration is in same direction as unbalanced force F is a vector; represents size & direction of a force F represents only the magnitude. Force and Motion Copyright © McGraw-Hill Education

4-1: Force Identify object(s) of interest = system Everything around the object that exerts forces on it is called the external world. Each force has a cause called the agent. A contact force exists when an object from the external world touches a system and thereby exerts a force on it. Field forces are exerted without contact. Force and Motion Copyright © McGraw-Hill Education

4-1: Force A physical representation that shows the forces acting on a system is called a free-body diagram. Force and Motion Copyright © McGraw-Hill Education

4-1: Force Force vectors in the same direction: combine them Combining Forces Force vectors in the same direction: combine them Forces in opposite directions: subtract resulting vector in the direction of the greater force. Vector sum of all the forces on an object is net force. Copyright © McGraw-Hill Education

4-1: Force Acceleration and Force As the applied force increases, so does the object’s acceleration. As the top graph shows, acceleration is proportional to the force. If the mass is increased, but the force is kept constant, the acceleration decreases. As the bottom graph shows, acceleration is inversely proportional to the mass. Copyright © McGraw-Hill Education Force and Motion

4-1: Force Newton’s Second Law Newton’s Second Law Newton’s second law states that the acceleration of an object is proportional to the net force acting on the object and inversely proportional to the mass of the object. To use Newton’s second law to solve a problem, follow these steps. Draw a free-body diagram showing the direction and relative strength of each force acting on the system. Then add the force vectors to find the net force. Next, use Newton’s second law to calculate the acceleration. Finally, if necessary, use what you know about accelerated motion to find the velocity or position of the object. Newton’s Second Law Copyright © McGraw-Hill Education Force and Motion

4-1: Force Newton’s Second Law ● 20.0 N 15.0 N Problem Response Use with Example Problem 1. Problem Kwaku pushes one end of a table with a force of 15.0 N. Salali pushes on the other end of the table with a force of 20.0 N. What is the net force on the table? SOLVE FOR THE UNKNOWN Determine the net force. Response SKETCH AND ANALYZE THE PROBLEM Draw the situation and a free-body diagram. Identify the knowns and unknowns. That is, Fnet = 5.0 in the direction Salali is pushing. KNOWN UNKNOWN FKwaku = 15.0 N Fnet = ? FSalali = −20.0 N EVALUATE THE ANSWER The net force is in the direction of the force with the large magnitude. Force and Motion Copyright © McGraw-Hill Education

4-1: Force Newton’s First Law An object that is at rest will remain at rest, and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on that object is zero Law of Inertia Inertia—tendency of an object to resist changes in velocity. If net force on object is zero, then the object is in equilibrium. If there is no net force acting on the object, then the object does not experience a change in speed or direction and is in equilibrium. Copyright © McGraw-Hill Education Force and Motion

4-2: Weight and Drag Force Weight is the gravitational force experienced by an object. (a field force) Measured in Newtons (N). Fg = mg or w = mg Near Earth’s surface, g is 9.8 N/kg toward Earth’s center. (= 9.8 m/s2) Weight and Drag Force Copyright © McGraw-Hill Education

4-2: Weight and Drag Force 35.0 kg ● 502 N Fg = mg Weight Use with Example Problem 2. Problem Arnold needs to lift a 35.0-kg rock. If he exerts an upward force of 502 N on the rock, what is the rock’s acceleration? SOLVE FOR THE UNKNOWN Determine the net force. Use Newton’s Second Law to find the acceleration. Response SKETCH AND ANALYZE THE PROBLEM Draw the situation & a free-body diagram. List the knowns and unknowns. KNOWN UNKNOWN mrock = 35.0 kg a = ? g = −9.8 m/s2 EVALUATE THE ANSWER The sign makes sense because the net force is upward. Copyright © McGraw-Hill Education Weight and Drag Force

4-2: Weight and Drag Force When you step on a scale, the scale exerts an upward support force on you. Apparent weight is the support force exerted on an object. Apparent weight depends on how you are accelerating. Weightlessness is the condition where there are no contact forces acting to support the object and the object’s apparent weight is zero. Copyright © McGraw-Hill Education Weight and Drag Force

4-2: Weight and Drag Force Use with Example Problem 3. Problem You are in an elevator, standing on a bathroom scale. You notice that the scale reads a weight that is less than your actual weight. (Assume that the scale is correctly calibrated.) a. Is the elevator moving at constant velocity, or is it accelerating? b. If it is accelerating, what is the direction of the acceleration? Response a. Since the support force does not equal your weight, you must be accelerating. b. The support force on you is less than your weight, so the elevator must be accelerating downward. Weight and Drag Force Copyright © McGraw-Hill Education

4-2: Weight and Drag Force A drag force is the force exerted by a fluid on an object opposing motion through the fluid. This force is dependent on: the motion of the object, the properties of the object, the properties of the fluid (viscosity and temperature) that the object is moving through. The constant velocity that is reached when the drag force equals the force of gravity is called the terminal velocity. Copyright © McGraw-Hill Education Weight and Drag Force

4-3: Newton’s Third Law Interaction Pairs When you exert a force on your friend to push him forward (FA on B), he exerts an equal and opposite force on you (FB on A), which causes you to move backward. The forces FA on B and FB on A are an interaction pair. An interaction pair is two forces that are in opposite directions, have equal magnitudes, and act on different objects. Newton’s third law states that all forces come in pairs and that the two forces are equal in strength, opposite in direction, and act on different objects. Newton’s Third Law Copyright © McGraw-Hill Education Newton’s Third Law

4-3: Newton’s Third Law Interaction Pairs When identifying an interaction pair, remember that they always occur in two different free-body diagrams and they always have the symmetry in subscripts. Copyright © McGraw-Hill Education Newton’s Third Law

4-3: Newton’s Third Law Interaction Pairs ● Problem ● FEarth on you ● Use with Example Problem 4. Problem You are walking along when you slip on some ice and fall. For a moment you are in free fall. During this time, what force do you exert on Earth if your mass is 55 kg? Fyou on Earth ● SOLVE FOR THE UNKNOWN Use Newton’s third law and the definition of weight to determine the force you exert on Earth. Response ANALYZE AND SKETCH THE PROBLEM Draw the situation and a free-body diagram. Identify the knowns and unknowns. KNOWN UNKNOWN myou = 55 kg Fyou on Earth = ? g = −9.8 N/kg EVALUATE THE ANSWER The force you exert on Earth is equal in magnitude and opposite in direction to the force Earth exerts on you. Newton’s Third Law Copyright © McGraw-Hill Education

4-3: Newton’s Third Law Tension The force exerted by a string or rope is called tension. At any point in a rope, the tension forces are pulling equally in both directions. Copyright © McGraw-Hill Education Newton’s Third Law

4-3: Newton’s Third Law Tension Problem ● Response 100 N Fg = mg Use with Example Problem 5. Problem You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 100 N, what is the maximum acceleration you can give the fish as you reel it in? SOLVE FOR THE UNKNOWN Use Newton’s Second Law to find the maximum acceleration. Response ANALYZE AND SKETCH THE PROBLEM Draw the situation and a free-body diagram. Identify the knowns and unknowns. KNOWN UNKNOWN mfish = 6 kg amax = ? FT, max = 100 N g = −9.8 N/kg EVALUATE THE ANSWER The magnitude of the acceleration is similar to the acceleration due to gravity, so it is reasonable. Newton’s Third Law Copyright © McGraw-Hill Education

4-3: Newton’s Third Law Normal Force The normal force is the perpendicular contact force exerted by a surface on another object. The normal force is not always equal to the object’s weight, as shown below. Finding the normal force is important when calculating the effects of friction. Copyright © McGraw-Hill Education Newton’s Third Law