Force Physics PDF

LESSON 7 FORCE  MECHANICS is the study of the motion of objects and the related concepts of force and energy. KINEMATICS DYNAMICS describes describes HOW OBJECTS MOVE WHY OBJECTS...

LESSON 7 FORCE  MECHANICS is the study of the motion of objects and the related concepts of force and energy. KINEMATICS DYNAMICS describes describes HOW OBJECTS MOVE WHY OBJECTS MOVE  KINEMATICS DYNAMICS describes describes HOW OBJECTS MOVE WHY OBJECTS MOVE MOTION MOTION + FORCE Displacement Force Time Momentum Velocity Energy Acceleration Work Mass Power  What is FORCE? PUSH OR PULL  Types of FORCE 1. CONTACT FORCE 2. NONCONTACT FORCE  CONTACT FORCE Applied Force Frictional Force Air resistance Normal Force NORMAL FORCE APPLIED FORCE (Push) AIR RESISTANCE FRICTIONAL FORCE  NONCONTACT FORCE MAGNETIC FORCE GRAVITATIONAL FORCE ELECTRIC FORCE  1 ST LAW OF MOTION NEWTON’S FIRST LAW OF MOTION: ( LAW OF INERTIA ) “Every object continues in its state of rest, or of uniform velocity in straight line, as long as no net force acts on it. “An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by unbalanced force”  1 ST LAW OF MOTION NEWTON’S FIRST LAW OF MOTION: What is INERTIA? ( LAW OF INERTIA ) ✓ Inertia is the tendency of an object to resist changes in its state of motion. The state of motion is defined by its velocity – the speed with direction. ✓ Thus, inertia could be redefined as: Inertia is the tendency of an object to resist changes in its velocity. ✓ Thus, we could provide an alternative means of defining inertia: Inertia is the tendency of an object to resist acceleration  1 ST LAW OF MOTION NEWTON’S FIRST LAW OF MOTION: “An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by unbalanced force” When do we say that the forces on the object are BALANCED and UNBALANCED?  1 ST LAW OF MOTION BALANCED FORCES NORMAL FORCE The ground is applying an equal amount of force exactly in the opposite direction. NET FORCE (perpendicular to the surface) The sum of all forces acted on an object NET FORCE = 0 GRAVITATIONAL FORCE Trying to pull the ball towards the center of the earth. The ball is STATIONARY (v = 0)  1 ST LAW OF MOTION BALANCED FORCES NORMAL FORCE The ground is applying an equal amount of force exactly in the opposite direction. NET FORCE (perpendicular to the surface) The sum of all forces acted on an object NET FORCE = 0 GRAVITATIONAL FORCE Trying to pull the ball towards the center of the earth. The ball is CONSTANTLY MOVING at frictionless surface. (v = 3 m/s)  1 ST LAW OF MOTION UNBALANCED FORCES NORMAL FORCE APPLIED FORCE (Push) FRICTIONAL FORCE Acts to resist the relative motion of an object The ball is STATIONARY The ball is MOVING (vi = 0) (vf = 3 m/s) NET FORCE ≠ 0 GRAVITATIONAL FORCE  1 ST LAW OF MOTION CONCEPTUAL EXAMPLE A school bus comes to a sudden stop, and all of the backpacks on the floor starts to slide forward. What force causes them to do that? Response: It isn’t the force that does it. By Newton’s first law, the backpacks continue their state of motion, maintain their velocity. The backpacks slow down if a force is applied, such as friction with the floor. 1 ST LAW OF MOTION 1 ST LAW OF MOTION  2 ND LAW OF MOTION NEWTON’S SECOND LAW OF MOTION: ( LAW OF ACCELERATION ) “The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to the object’s mass. The direction of the acceleration is in the direction of the net force acting on the object”  2 ND LAW OF MOTION FORCES ARE UNBALANCED ACCELERATION NET FORCE MASS  2 ND LAW OF MOTION NEWTON’S SECOND LAW OF MOTION: ( LAW OF ACCELERATION ) “The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to the object’s mass. The direction of the acceleration is in the direction of the net force acting on the object”. Newton’s second law of motion can be written as an equation: Where a stands for acceleration, m for the mass, and σ𝐅 σ𝐅 for the net force on the object. The symbol σ (Greek 𝒂= “sigma”) stands for “the sum of”; F stands for force so 𝑚 σ𝐅 means the vector sum of all forces acting on the object, which we define as net force.  2 ND LAW OF MOTION APPLIED FORCE 20 UNITS MORE ACCELERATION LESS MASS APPLIED FORCE 20 UNITS LESS ACCELERATION MORE MASS  2 ND LAW OF MOTION NEWTON’S SECOND LAW OF MOTION: ( LAW OF ACCELERATION ) σ𝐅 = ma m = 1.0 kg = 1.0 kg x 1.0 m/s2 = 1kg·m/s2 a = 1.0 m/s2 1 N = 1kg·m/s 2  2 ND LAW OF MOTION MASS AND WEIGHT MASS is the quantity of inertia possessed by an object. Amount of substance in an object. WEIGHT A gravitational force that attracts or pulls any object towards its origin. Weight = mass x acceleration due to gravity W = mg  2 ND LAW OF MOTION MASS AND WEIGHT Weight of the person on Earth W = mg W = (25 kg) (9.8 m/s ) 2 W = 245 kg-m/s2 W = 245 N Weight of the person on Moon W = mg W = (25 kg) (1.6 m/s2) W = 40 kg-m/s2 W = 40 N  SAMPLE PROBLEM 1 FORCE TO ACCELERATE A FAST CAR Estimate the net force needed to accelerate (a) a 1000-kg car at 5m/s2; (b) a 200-g apple at the same rate. a. a 1000-kg car at 5m/s 2 Given: Solution: m = 1000 kg σF = ma a = 5 m/s2 σF = (1 000 kg) (5 m/s2) Required: σF = 5 000 kg-m/s2 σ𝐅 (net force) σF = 5 000 N Equation: Answer: σ𝐅 = ma 5 000.00 N, East  SAMPLE PROBLEM 1 FORCE TO ACCELERATE A FAST CAR Estimate the net force needed to accelerate (a) a 1000-kg car at 5m/s2; (b) a 200-g apple at the same rate. b. a 200-g apple at 5 m/s2 Solution: Given: σF = ma m = 200 g (0.2 kg) σF = (0.2 kg) (5 m/s2) a = 5 m/s2 Required: σF = 1 kg-m/s2 σ𝐅 (net force) σF = 1 N Equation: Answer: σ𝐅 = ma 1.00 N, East  SAMPLE PROBLEM 2 FORCE TO STOP A CAR What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h (27.8 m/s) within a distance of 55 m? m = 1500 kg vix = 27.8 m/s vfx = 0 m/s dix = 0 m dfx = 55 m  SAMPLE PROBLEM 2 FORCE TO STOP A CAR What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h (27.8 m/s) within a distance of 55 m? Given: m = 1500 kg σF = ma vix = 100 km/h (27.8 m/s) vfx = 0 m/s vfx = vix + at dx = (55 m – 0 m) = 55 m vix + vfx dx = ( )t Required: 2 1 σF (net force) dx = vixt + at² 2 Equation: vfx² = vix² + 2axdx vfx² = vix² + 2axdx σF = ma  SAMPLE PROBLEM 2 FORCE TO STOP A CAR What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h (27.8 m/s) within a distance of 55 m? Given: m = 1500 kg vfx = vix + 2axdx 2 2 vix = 100 km/h (27.8 m/s) vfx2 - vix2 = 2axdx vfx = 0 m/s 2 vfx 2 −vix 2axdx = dx = (55 m – 0 m) = 55 m 2dx 2dx 2 2 Required: vfx − vix = ax σF (net force) 2dx Equation: 2 2 vfx −vix vfx² = vix² + 2axdx ax = 2dx σF = ma  SAMPLE PROBLEM 2 FORCE TO STOP A CAR What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h (27.8 m/s) within a distance of 55 m Given: Solution: m = 1500 kg v2fx −v2ix ax = vix = 100 km/h (27.8 m/s) 2dx vfx = 0 m/s (0)2 −(27.8 m/s)2 a= 2(55 m) dx = (55 m – 0 m) = 55 m − 772.84m2 /s)2 Required: a= 110 m σF (net force) 𝐚 = −𝟕. 𝟎𝟑 m/s2 Equation: Answer: vfx² = vix² + 2axdx a = 𝟕. 𝟎𝟑 m/s2, West σF = ma  SAMPLE PROBLEM 2 FORCE TO STOP A CAR What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h (27.8 m/s) within a distance of 55 m? Given: Solution: m = 1500 kg σF = ma vix = 100 km/h (27.8 m/s) σF = (1500 kg)(- 7.03 m/s2) vfx = 0 m/s σF = - 10,545 kg- m/s2 dx = (55 m – 0 m) = 55 m σF = - 10,545 N ax = - 7.03 m/s2 Required: σF (net force) Answer: Equation: 10,545.00 N, West σF = ma  rd 3 LAW OF MOTION NEWTON’S THIRD LAW OF MOTION: ( LAW OF INTERACTION ) “Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first”.  rd 3 LAW OF MOTION NEWTON’S THIRD LAW OF MOTION: ( LAW OF INTERACTION ) Force exerted on the cart Force exerted on the person by the man by the cart  rd 3 LAW OF MOTION NEWTON’S THIRD LAW OF MOTION: ( LAW OF INTERACTION ) Force exerted on the hammer by the nail Force exerted on the nail by the hammer  rd 3 LAW OF MOTION NEWTON’S THIRD LAW OF MOTION: ( LAW OF INTERACTION )  UNIVERSAL LAW OF GRAVITATION NEWTON’S LAW UNIVERSAL GRAVITATION “Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles.”  UNIVERSAL LAW OF GRAVITATION NEWTON’S LAW UNIVERSAL GRAVITATION “Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles.” Where m1 and m2 are the masses of the two m1 m2 Fg = G 2 particles, r is the distance between them, G is a universal constant which must be measured r experimentally. G = 6.67 x 10 -11 N-m /kg 2 2  SAMPLE PROBLEM 3 m1 = 50 kg m2 = 70 kg Can you attract another person gravitationally? A 50-kg person and a 70-kg person are sitting on a bench close to each other. Estimate the magnitude of the gravitational force each exerts on the other APPROACH: This is an estimate: we let the distance between the centers of the two people be ½ m (about as close as you can get). r = 0.5 m  SAMPLE PROBLEM 3 m1 = 50 kg m2 = 70 kg Given: m1 = 50 kg m2 = 70 kg r = 0.50 m G = 6.67 x 10-11 N-m2/kg2 Required: Fg (Gravitational force) Equation: m1 m2 Fg = G 2 r r = 0.5 m  SAMPLE PROBLEM 3 Given: Solution: m1 = 50 kg Fg = m1 m2 G 2 r m2 = 70 kg − N⋅m2 r = 0.50 m ( 6.67 x 10 11 kg2 )(50 kg)(70 kg) Fg = G = 6.67 x 10-11 N-m2/kg2 (0.50 m)2 Required: ( 2.335 x 10−7 N⋅m2 ) Fg = Fg (Gravitational force) (0.25 m2 ) Fg = 9.34 x 10-7 N Equation: m1 m2 Fg = G 2 Answer: r 9.34 x 10-7 N  SAMPLE PROBLEM 4 Determine the force of gravitational attraction between Earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40,000 ft above Earth’s surface. This would place the 6 students a distance of 6.39 x 10 m from Earth center.  SAMPLE PROBLEM 4 Given: Solution: m1 m2 m1 = 5.98 x 10 kg (Earth’s mass) 24 Fg = G 2 r m2 = 70 kg (student’s mass) 2 N⋅m r = 6.39 x 10 m 6 ( 6.67 x 10− 11 kg2 )(5.98 x 1024 kg)(70 kg) G = 6.67 x 10-11 N-m2/kg2 Fg = (6.39 x 106 m)2 ( 2.792 x 1016 N⋅m2 ) Required: Fg = (4.083 x 1013 m2 ) Fg (Gravitational force) Fg = 683.81 N Equation: Answer: m1 m2 Fg = G 2 683.81 N r