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22 May 2017 Norah Ali Al- moneef 1 3 .1 Force,weight,and gravitational mass What is a Force? Force is an action that can change motion. A force is what we call a push or a pull, or any action that has the ability to change an object’s motion. •the force causes a change in velocity, or an acceleration • Because velocity is a vector, having both magnitude and direction, a force can alter an objects speed, its direction, or both. • All forces have a strength (magnitude) and a direction. Therefore • , forces are vector quantities. •Forces are either balanced or unbalanced •There are two types of forces that we can see •Contact Forces •At a distance forces 22 May 2017 Norah Ali Al- moneef Fkick Fgrav 2 Contact Force Forces in which the two interacting objects are physically in contact with each other. EXAMPLES •Hitting •Pulling with a rope •Lifting weights •Pushing a couch •Frictional Force •Tensional Force •Normal Force •Air Resistance Force •Applied Force •Spring Force 22 May 2017 Norah Ali Al- moneef 3 At A Distance Force (field force) •Forces in which the two interacting objects are not in physical contact with each other, but are able to exert a push or pull despite the physical separation. Example: •Gravitational Force •Electrical Force •Magnetic Force •Nuclear Forces 22 May 2017 Norah Ali Al- moneef 4 22 May 2017 Norah Ali Al- moneef 5 Measuring Forces • Forces are measured in newtons (kg . m/s2). • Forces are measured using a spring scale. 22 May 2017 Norah Ali Al- moneef 6 Balanced Force • A force that produces no change in an object’s motion because it is balanced by an equal, opposite force. • no change in velocity The object shown in the diagram at rest since there is no net force acting on it. If you were to add these forces they would = 0 FALSE! A net force does not cause motion. A net force causes a change in motion, or acceleration. 7 Norah Ali Al- moneef 22 May 2017 Unbalanced Forces • Are forces that results in an object’s motion being changed. + • velocity changes (object accelerates) Fnet Fpull Ffriction N 22 May 2017 W N Norah Ali Al- moneef 8 22 May 2017 Norah Ali Al- moneef 9 example On a horizontal, frictionless surface, the blocks above are being acted upon by two opposing horizontal forces, as shown. What is the magnitude of the net force acting on the 3kg block? A. zero B. 2N C. 1.5 N D. 1N E. More information is needed. 22 May 2017 Norah Ali Al- moneef 10 example • A woman is holding two dogs on a leash. • If each dog pulls with a force of 80 N, how much force does the woman have to exert to keep the dogs from moving? 160 N to the left example 22 May 2017 Norah Ali Al- moneef 11 example In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view of the Fig. • The tire remainsstationary in spite of the three pulls. Alex pulls with force FA of magnitude 220 N, and Charles pulls with force FC of magnitude 170 N. The direction of FC is not given. What is the magnitude of Betty's force FB ? 22 May 2017 Norah Ali Al- moneef 12 FA FB FC m(0) 0 FB FA FC FBy FAy FCy FB sin ( 90 ) FA sin 133 FC sin , FB ( 220 N ) (sin 133 ) (170 N ) sin Force components in the x-axis : FBX FAX FCX FB cos ( 90 ) FA cos 133 FC cos 0 ( 220 N) (cos 133 ) (170 N) cos cos 1 (220 N) (cos 133 ) 28.04 170 N Therefore, FB 241 N 22 May 2017 Norah Ali Al- moneef 13 Weight Weight = Mass x Gravity W m g The weight of an object is the gravitational force that the planet exerts on the object. The weight always acts downward, toward the center of the planet. SI Unit of Weight: : newton (N) – m: mass of the body (units: kg) – g: gravitational acceleration (9.8m/s2, • As the mass of a body increases, its’ weight increases proportionally 22 May 2017 Norah Ali Al- moneef 14 • Weight – the force of gravity on an object W = mg MASS 22 May 2017 always the same Norah Ali Al- moneef (kg) W: weight (N) m: mass (kg) g: acceleration due to gravity (m/s2) WEIGHT depends on gravity (N) 15 Mass … • is a measure of how much matter there is in something - usually measured in kilograms in science • causes an object to have weight in a gravitational field •A measure of the resistance of an object to changes in its motion due to a force (describes how difficult it is to get an object moving) •Scalar Note that mass is involved in the force of gravity! This is a separate property from that of inertia, so we give this property the name gravitational mass. 22 May 2017 Norah Ali Al- moneef 16 Example : • what is the weight of a 2 kg mass? • W = Fg = mxg = 2 kg x 9.8 m/s2 = 19.6 N Example : • What is the mass of a 1000 N person? • W = Fg = mxg m = Fg/g = 1000 N / 9.8 m/s2 =102 kg Example : • A girl weighs 745 N. What is his mass m=F÷g m = (745 N) ÷ (9.8 m/s2) m = 76.0 kg 22 May 2017 Norah Ali Al- moneef 17 3.2 Density • A measure of how much mass is in a certain volume of space • The density of a substance is equal to its mass divided by its volume Relative Density •The density of a material or substance, relative to another substance •Water is the substance to which we generally compare other substances •ALSO known as SPECIFIC GRAVITY •For water: Absolute Density = 1000kg/m3 22 May 2017 Norah Ali Al- moneef 18 Example : 22 May 2017 Norah Ali Al- moneef 19 Example : 22 May 2017 Norah Ali Al- moneef 20 Ex -3-17 Find the density of gasoline if 5 Kg has a volume of 7.35 x10 -3 m 3 And its relative density (specific gravity) Relative density =( 0.68 x 10 3 Kg / m 3) / 10 3 Kg / m 3 = 0.68 22 May 2017 Norah Ali Al- moneef 21 3.3 Newton's first law Every object continues in a state of rest , or of uniform motion in a straight line , unless it is compelled to change that state by forces acting upon it. An equivalent statement of the first law is that : An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force. This, at first, does not seem obvious. Most things on earth tend to slow down and stop. However, when we consider the situation, we see that there are lots of forces tending to slow the objects down such as friction and air resistance. 22 May 2017 Norah Ali Al- moneef 22 Newton’s First Law • Newton’s First Law of Motion – “Law of Inertia” • Inertia – tendency of an object to resist any change in its motion – increases as mass increases 22 May 2017 Norah Ali Al- moneef 23 Mass = Inertia – an object’s resistance to motion Would it be more difficult to pull an elephant or a mouse? 22 May 2017 Norah Ali Al- moneef 24 The net force acting on an object is the vector sum of all the forces acting on it. Examples: 9N 8N 8N 6N 4N 7N 12 N 8N ? 5N 4N 3N 7N 4N If an object is remaining at rest, it is incorrect to assume that there are no forces acting on the object. We can only conclude that the net force on the object is zero. 22 May 2017 Norah Ali Al- moneef 25 • The condition of zero acceleration is called equilibrium. • In equilibrium, all forces cancel out leaving zero net force. • Objects that are standing still are in equilibrium because their acceleration is zero. • Objects that are moving at constant speed and direction are also in equilibrium. • A static problem usually means there is no motion. 22 May 2017 Norah Ali Al- moneef 26 3.5Newton’s Third Law of Motion There is one further important aspect of motion that Newton identified: the distinction between forces that act on an object and forces that act by the object. This leads to his Third Law of Motion: For every force by a first object on a second object, there is a force by the second object on the first object with the same magnitude but in the opposite direction. This is sometimes called the law of action and reaction. I like to call it: you can’t push yourself! You can only push on an object and hope that it pushes back. Example: when you walk up a stairs, you use your muscles to push down on the stairs and you trust that the stairs will push back up on you lifting you up the stairs. 22 May 2017 Norah Ali Al- moneef 27 Newton’s Third Law • Newton’s Third Law of Motion – When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first. F AB F BA 22 May 2017 Norah Ali Al- moneef 28 • The action and reaction forces are EXACTLY 180o apart in direction. how come they don’t always cancel since they push on different objects., making net force and acceleration impossible? 22 May 2017 Norah Ali Al- moneef 29 Newton’s Third Law r F12 F21 Force on “1” due to “2” • Single isolated force cannot exist • For every action there is an equal and opposite reaction • “For every action force, there is … a reaction force” means: – Forces ALWAYS occur in pairs. – Single forces NEVER happen. 22 May 2017 Norah Ali Al- moneef 30 Newton’s Third Law • example: How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force? Aren’t these “balanced forces” resulting in no acceleration? Explanation: NO!!! – forces are equal and opposite but act on different objects – they are not “balanced forces” – the movement of the horse depends on the forces acting on the horse 22 May 2017 Norah Ali Al- moneef 31 Newton’s Third Law • Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left. Flying gracefully through the air, birds depend on Newton’s third law of motion. As the birds push down on the air with their wings, the air pushes their wings up and gives them lift. 22 May 2017 Norah Ali Al- moneef 32 Newton’s Third Law • Action-Reaction Pairs The rocket exerts a downward force on the exhaust gases. The gases exert an equal but opposite upward force on the FG rocket. FR 22 May 2017 According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. Norah Ali Al- moneef 33 Other examples of Newton’s Third Law • The baseball forces the bat to the left (an action); the bat forces the ball to the right (the reaction). • Locomotion is the act of moving or the ability to move from one place to another. 22 May 2017 Norah Ali Al- moneef 34 What happens if you are standing on a skateboard or a slippery floor and push against a wall? You slide in the opposite direction (away from the wall), because you pushed on the wall but the wall pushed back on you with equal and opposite force. Why does it hurt so much when you stub your toe? When your toe exerts a force on a rock, the rock exerts an equal force back on your toe. The harder you hit your toe against it, the more force the rock exerts back on your toe (and the more your toe hurts). 22 May 2017 Norah Ali Al- moneef 35 Calculate force • Three people are each applying 250 N of force to try to move a heavy cart. • The people are standing on a rug. • Someone nearby notices that the rug is slipping. • How much force must be applied to the rug to keep it from slipping? • Sketch the action and reaction forces acting between the people and the cart and between the people and the rug. The third law says that each of the forces applied creates a reaction force. So Each person applies a force to the cart and the cart applies an equal and opposite force to the person. The force on the rug is the sum of the reaction forces acting on each person. The total force that must be applied to the rug is 750 N in order to equal the reaction forces from all three people. 22 May 2017 Norah Ali Al- moneef 36 • Locomotion is the act of moving or the ability to move from one place to another. 22 May 2017 Norah Ali Al- moneef 37 Action-eaction Pairs r n n Fg Fg' Define the OBJECT (free body) • Newton’s Law uses the forces acting ON object • N and Fg act on object • n’ and Fg’ act on other objects 22 May 2017 Norah Ali Al- moneef 38 Normal Force Is Not Always Equal to the Weight 22 May 2017 Norah Ali Al- moneef 39 Normal Force The normal force is the force exerted by a surface on an object. and always perpendicular to the surface that produces it. 22 May 2017 Norah Ali Al- moneef 40 Normal Force The normal force may be equal to, greater than, or less than the weight. 22 May 2017 Norah Ali Al- moneef 41 The Other Logical Problem • If the Newton’s Third Law action and reaction forces are always equal and opposite, how do two objects of different sizes get different accelerations in the same interaction? (When a bug hits a windshield, different things happen to the bug and windshield.) • If the action and reaction forces are the same size, how can two objects push on each other and get different accelerations? • Newton’s Second Law says that the acceleration of an object depends not only on the force on it, but on the object’s mass. 22 May 2017 Norah Ali Al- moneef 42 3.6 Newton’s Second Law • Newton’s Second Law of Motion • The force F needed to produce an acceleration a is F = ma – The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. • The same force acting on objects of different mass will produce different accelerations! Same Force Fnet m 22 May 2017 a Fnet = Norah Ali Al- moneef m =a 43 Newton’s Second Law • The acceleration of an object is equal to the force you apply divided by the mass of the object. •Force and acceleration are vectors . They both have: Magnitude and direction if you exert a force east . The object will accelerate east 22 May 2017 Norah Ali Al- moneef 44 • The acceleration of an object is equal to the force you apply divided by the mass of the object. Newton’s Third Law • Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the mass of the object. a F m Small mass more acceleration Large mass less acceleration 22 May 2017 Norah Ali Al- moneef 45 • If you apply more force to an object, it accelerates at a higher rate. • If an object has more mass it accelerates at a lower rate because mass has inertia. 22 May 2017 Norah Ali Al- moneef 46 22 May 2017 Norah Ali Al- moneef 47 Newton’s second Law F =ma a F m F m a m: mass (kg) a:acceleration (m/s2) 1 N = 1 kg ·m/s2 Three forms of the second law: 22 May 2017 Norah Ali Al- moneef 48 22 May 2017 Norah Ali Al- moneef 49 More about F = ma If you double the mass, you double the force. If you double the acceleration, you double the force. What if you double the mass and the acceleration? (2m)(2a) = 4F Doubling the mass and the acceleration quadruples the force. So . . . what if you decrease the mass by half? How much force would the object have now? 22 May 2017 Norah Ali Al- moneef 50 Example : A railway engine pulls a wagon of mass 10 000 kg along a straight track at a steady speed. The pull force in the couplings between the engine and wagon is 1000 N. a.What is the force opposing the motion of the wagon? b.If the pull force is increased to 1200 N and the resistance to movement of the wagon remains constant, what would be the acceleration of the wagon? Solution a) When the speed is steady, by Newton’s first law, the resultant force must be zero. The pull on the wagon must equal the resistance to motion. So the force resisting motion is 1000 N. b) The resultant force on the wagon is 1200 – 1000 = 200 N By Equation 22 May 2017 Norah Ali Al- moneef 51 Example : a) Find the acceleration of a 20 kg crate along a horizontal floor when it is pushed with a resultant force of 10 N parallel to the floor. b) How far will the crate move in 5s (starting from rest)? Solution 22 May 2017 Norah Ali Al- moneef 52 Example : A 1kg stone fall freely from rest from a bridge. a -What is the force causing it to accelerate? b -What is its speed 4s later? c -How far has it fallen in this time? Solution The force causing it to fall is its weight. As it is falling with acceleration due to gravity b) c) 22 May 2017 Norah Ali Al- moneef 53 Equilibrium • When the net force acting on an object is zero, the forces on the object are balanced. • We call this condition equilibrium. 22 May 2017 Norah Ali Al- moneef 54 Equilibrium • A moving object continues to move with the same speed and direction. • Newton’s second law states that for an object to be in equilibrium, the net force, or the sum of the forces, has to be zero. • Acceleration results from a net force that is not equal to zero. 22 May 2017 Norah Ali Al- moneef 55 22 May 2017 Norah Ali Al- moneef 56 Example : 22 May 2017 Norah Ali Al- moneef 57 Balancing Forces (Statics) There are often situations where a number of forces are acting on something, and the object has no motion – it is STATIC or in EQUILIBRIUM or at rest. • This means the NET FORCE on the object is zero, or in other words the forces balance each other out. • •ΣF = m a a =0 ΣF = 0 Objects in equilibrium (no net external force) also move at constant velocity In a two-dimensional problem, we can separate this equation into its two component, in the x and y directions: Σ Fx = 0 Σ Fy = 0 unbalanced Forces (Dynamics) •There are other situations where all the forces acting on something do not cancel each other out completely. •This means the NET FORCE on the object is not zero, the object will change its motion and accelerate proportional to the object’s mass. •ΣF = m a 22 May 2017 Norah Ali Al- moneef 58 Example : Calculating the net force from four forces Four people are pulling on the same 200 kg box with the forces shown. Calculate the acceleration of the box. 1. Use a = F ÷ m. 2. First add the forces to find the net force. F = - 75N - 25N + 45N + 55N = 0 N, a=0 22 May 2017 Norah Ali Al- moneef 59 Example : Using equilibrium to find an unknown force Two chains are used to lift a small boat. One of the chains has a force of 600 N. Find the force on the other chain if the mass of the boat is 150 kilograms. 1. 2. 3. Use: net force = zero, Fw = mg and g = 9.8 N/kg. Fw = mg = (150 kg)(9.8 N/kg) = 1,470 N. Let F be the force in the other chain, equilibrium requires: – F + (600 N) = 1,470 N F = 1,470 N – 600 N F = 870 N. 22 May 2017 Norah Ali Al- moneef 60 Applications of equilibrium • Real objects can move in three directions: up-down, right-left, and front-back. • The three directions are called three dimensions and usually given the names x, y, and z. • When an object is in equilibrium, forces must balance separately in each of the x, y, and z dimensions. 22 May 2017 Norah Ali Al- moneef 61 example • Three people are pulling on a wagon applying forces of 100 N,150 N, and 200 N. • The wagon has a mass of 25 kilograms. • Determine the acceleration and the direction the wagon moves. a =200 –(100+150) / 25 = - 50 / 25 = - 2 m / s2 22 May 2017 Norah Ali Al- moneef 62 Example : What force would be required to accelerate a 40 kg mass by 4 m/s2? F = ma F = (40 kg)(4 m/s2) F = 160 N Example : A 4.0 kg shotput is thrown with 30 N of force. What is its acceleration? a=F÷m a = (30 N) ÷ (4.0 kg) a = 7.5 m/s2 22 May 2017 Norah Ali Al- moneef 63 Example : 22 May 2017 Norah Ali Al- moneef 64 Example A dragster accelerates from 0 to 60 m/s in 6 seconds. Calculate the force on the car if the mass is 103 kg. F = ma and a = ∆v / ∆t Answer: 104 N 22 May 2017 Norah Ali Al- moneef 65 Solving Force Problems A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000 m/s2. (a) What force acts on the bullet? (b) What force acts on the rifle? ( c) If the bullet is in the rifle for 0.007 seconds, what is its muzzle velocity? a = 30,000 m/s2 (a) F = ma, F = 0.009 kg x 30,000 m/s2 = 270 N(grams must be converted to kilograms in order to use the MKS system and get newtons as force unit answer) (b) Newton’s 3rd Law (Action / Reaction) – the rifle is “shot” in the opposite direction with the same force as the bullet or – 270 N (c) (muzzle velocity means the velocity at which the bullet leaves the rifle barrel) Using the equation V = Vo + at, V = o + 30,000 x 0.007 = 210 m/s (Vo = 0 since the bullet is not moving before it is fired) 22 May 2017 Norah Ali Al- moneef 66 Solving Force Problems (a) How much force is needed to reduce the velocity of a 6400 Kg truck from 20 m/s to 10 m/s in 5 seconds? (b) What is its stopping distance? time = 5 sec 10 mph 20 mph • (a) F = ma, we must first find mass. The weight is 6400 Kg. Wt = mass x gravity therefore, m = w/g, mass = 6400 Kg /9.8 m/s2 = 653.06 Kg. • Now, to find acceleration, V = Vo + at or a = (V – Vo) / t • a = (20 – 10)/ 5 = - 2 m/s2 and F = 200 x (-2) = - 400 N (negative means the force is opposing the motion) • (b) Δx = Vot + ½ at2 we get Δx = (20 x 5) + ½ (-2) 52 = 75 m is the distance traveled during stopping. 22 May 2017 Norah Ali Al- moneef 67 Example : A force which gives a 2.0 kg object an acceleration of 1.6 m/s2 would give an 8.0 kg object what acceleration ? (A) 0.2 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 6.4 m/s2 Example : A car slows form 50 m/s to 15 m/s in 10 seconds when the brakes exert a force of 200 N. What is the weight of the car ? (A) 57 N (B) 560 N (C) 1830 lbs (D) 22,400N Example : A cable supports an elevator which is 2000 Kg. The tension in the cable lifting the elevator is 25 KN. What is the acceleration ? (A) 2.7 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 8.4 m/s2 22 May 2017 Norah Ali Al- moneef 68 Example : m = 1850 kg With two guys pushing…….what are The net forces??? F + 275 N + 395 N 560 N = +110 N a 22 May 2017 F m 110 N 0.059 m/s 2 1850 kg Norah Ali Al- moneef 69 If the airplane’s mass is 13 300 kg, what is the magnitude of the net force that the catapult and jet engine exert on the plane? 2 F ma (13 300 kg)(31 m/s ) = 4.1105 N 22 May 2017 Norah Ali Al- moneef 70 Where does the calculus fit in? dv d 2x F ma m m dt dt There could be situations where you are given a displacement function or velocity function. The derivative will need to be taken once or twice in order to get the acceleration. Here is an example. You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72-kg. The elevator starts from rest and travels upward with a speed that varies with time according to: 2 v(t ) 3t 0.20t When t = 4.0s , what is the reading on the bathroom scale (a.k.a. Force Normal)? dv d (3t 0.20t ) a 3 0.40t dt dt a (4) 3 0.40(4) 4.6 m / s2 2 22 May 2017 Fnet ma N mg ma N ma mg N (72)(9.8) (72)( 4.6) 1036.8 N Norah Ali Al- moneef 71 Example on Newton’s 2. and 3. Law Suppose you are an astronaut in outer space giving a brief push to a spacecraft whose mass is bigger than your own 1) Compare the magnitude of the force you exert on the spacecraft, FS, to the magnitude of the force exerted by the spacecraft on you, FA, while you are pushing: 1. FA = FS 2. FA > FS 3. FA < FS correct Third Law! 2) Compare the magnitudes of the acceleration you experience, aA, to the magnitude of the acceleration of the spacecraft, aS, while you are pushing: 1. aA = aS a=F/m 2. aA > aS correct F same lower mass gives larger a 3. aA < aS 22 May 2017 Norah Ali Al- moneef 72 Example: You are stranded in space, away from your spaceship. Fortunately, you have a propulsion unit that provides a constant net force F for 3.0 s. You turn it on, and after 3.0 s you have moved 2.25 m. If your mass is 68 kg, find F. x v0t 12 at 2 12 at 2 2x 2(2.25 m) 2 a 2 0.50 m/s t (3.0 s)2 a axiˆ 0.50 m/s2iˆ F ma (68 kg)(0.50 m/s 2 )iˆ 34 Niˆ 22 May 2017 Norah Ali Al- moneef 73 Example: Three Forces Moe, Larry, and Curley push on a 752 kg boat, each exerting a 80.5 N force parallel to the dock. (a) What is the acceleration of the boat if they all push in the same direction? (b) What is the acceleration if Moe pushes in the opposite direction from Larry and Curley as shown? F1 FM FL FC 3(80.5 N) 241.5 N a1 F1 / m (241.5 N) / (752 kg) 0.321 N/kg 0.321 m/s 2 F2 FM FL FC 80.5 N 22 May 2017 a2 F2 / m (80.5 N) / (752 kg) 0.107 m/s 2 Norah Ali Al- moneef 74 Newton’s Second Law of Motion Note that this is a vector equation, and should really be worked in component form: S Fx = max S Fy = may . We can now see that Newton’s First Law of Motion is really just a special case of his Second Law of Motion. 22 May 2017 Norah Ali Al- moneef 75 example In the Figs. one or two forces act on a puck that moves over frictionless ice and along an x axis, in one-dimensional motion. • The puck's mass is m = 0.20 kg. Forces F 1 and F 2 are directed along the axis and have magnitudes F1 = 4.0 N and F2 = 2.0 N. Force F 3 is directed at angle = 30° and has magnitude F3 = 1.0 N. In each situation, what is the acceleration of the puck? F ma . net , x x F1 4.0 N 20 m / s 2 . m 0.20 kg F F 4.0 N 2.0 N ax 1 2 10 m / s 2 . m 0.20 kg ax F3, x F2 F3 cos F2 m m (1.0 N) ( cos 30 ) 2.0 N 5.7 m / s 2 0.20 kg ax 22 May 2017 Norah Ali Al- moneef 76 example Solving Problems with Newton’s Laws: Free-Body Problems • Newton’s second law tells us that the acceleration of an object is proportional to the net force acting on an object. • The net force is the vector sum of all the forces acting on the object. FR = 141 N 22 May 2017 Norah Ali Al- moneef 77 example Adding force vectors. Calculate the sum of the two forces shown acting on the boat. 22 May 2017 Norah Ali Al- moneef 78 Example : a=3.02 m / s 22 May 2017 Norah Ali Al- moneef 79 Example : 22 May 2017 Norah Ali Al- moneef 80 22 May 2017 Norah Ali Al- moneef 81 Example : 22 May 2017 Norah Ali Al- moneef 82 Free-body diagrams • To keep track of the number and direction of all the forces in a system, it is useful to draw a free-body diagram. • A free-body diagram makes it possible to focus on all forces and where they act 22 May 2017 Norah Ali Al- moneef 83 Free-body diagrams • Forces due to weight or acceleration may be assumed to act directly on an object, often at its center. • A reaction force is usually present at any point an object is in contact with another object or the floor. • If a force comes out negative, it means the opposes another force. 22 May 2017 Norah Ali Al- moneef 84 Applications of equilibrium What is the upward force in each cable? 22 May 2017 • If an object is not moving, then you know it is in equilibrium and the net force must be zero. • You know the total upward force from the cables must equal the downward force of the sign’s weight because the sign is in equilibrium. Norah Ali Al- moneef 85 Tension force The force exerted by the string, at either end: • direction is parallel to the string • magnitude (same at both ends) is called the tension Example. Suppose a string can withstand string tension 500 N without breaking. What is the maximum mass M that it can hold suspended in Earth’s gravity? Weight W = Mg. Static equilibrium implies T = W. So m = W/g = T/g (500 N) / (9.8 m/s^2) = 51.0 kg. The forces on either mass are weight and string tension. But the net force is 0; the two forces cancel 22 May 2017 Norah Ali Al- moneef 86 Rules for Ropes and Pulleys • Force from rope points AWAY from object – (Rope can only pull) • Magnitude of the force is Tension – Tension is same everywhere in the rope • Tension does not change when going over pulley Approximations: Neglect mass of rope and pulley, neglect friction in pulley 22 May 2017 Norah Ali Al- moneef 87 example: to ceiling box hangs from a rope attached SFy = may y T T - W = may T = W + may W In this case ay = 0 So T = W 22 May 2017 Norah Ali Al- moneef 88 The tension force • • • • The tension force pulls on a body. The cord is often assumed to be massless. We usually assume pulleys to be frictionless A string has a single tension force (magnitude). The direction depends on the body on which this force acts upon. The tension forces on two sides of a frictionless pulley are the same in magnitude. 22 May 2017 Norah Ali Al- moneef 89 3 . 7 the significance of Newton's laws of motion 3 .7 some examples of Newton's laws Make a diagram (conceptualize) Categorize: no acceleration (at rest ) F 0 F ma Isolate each object and draw a free body diagram for each object. Draw in all forces that act on the object. Establish a convenient coordinate system. Write Newton’s law for each body and each coordinate component. set of equations. Finalize by checking answers. F 0 accelerating object: 22 May 2017 Norah Ali Al- moneef 90 example the Figure shows a block S (the sliding block) with mass M = 3.3 kg. The block is free to move along a horizontal frictionless surface such as an air table. Find (a) the acceleration of the sliding block, (b) the acceleration of the hanging block, and (c) the tension in the cord. Consider mass M : The positive direction of a is to the right. Consider mass m : S F Ma T Ma T mg - ma m g 3.8 m / s 2 Mm Same acceleration for both (they move Mm T g 13 N together) Mm a 22 May 2017 Norah Ali Al- moneef 91 example • In Fig. a, a block B of mass M = 15.0 kg hangs by a cord from a knot K of mass mK, which hangs from a ceiling by means of two other cords. The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravitational force on the block. What are the tensions in the three cords? 22 May 2017 Norah Ali Al- moneef 92 At a knot, the tensions are different. The tensions on both sides of a frictionless pulley are the same. T3 Mg M (0) 0 T3 147 N At the knot, T1 T2 T3 0 T1 cos 28 T2 cos 47 0 0 T1 sin 28 T2 sin 47 147 N 0 T1 104 N and T2 134 N The tensions are 104 N in cord 1, 134 N in cord 2, and 147 N in cord 3 (b) We now cut the cord. As the block then slides down the inclined plane, does it accelerate? If so, what is its acceleration? mg sin ma Positive direction is down the inclined plane a g sin a (9.8 m / s 2 ) (sin 27 ) 4.4 m / s 2 22 May 2017 Norah Ali Al- moneef 93 example • In Fig. a, a constant horizontal force Fap of magnitude 20 N is applied to block A of mass mA = 4.0 kg, which pushes against block B of mass mB = 6.0 kg. The blocks slide over a frictionless surface, along an x axis. (a) What is the acceleration of the blocks? Fap FAB m A a Fap (m A m B )a Fap 20 N a 2.0 m / s 2 m A m B 4.0 kg 6.0 kg The acceleration of block A and block B is the same (b) What is the force FBA on block B from block A (Fig. c)? FBA mB a FBA (6.0 kg) (2.0 m / s 2 ) 12 N 22 May 2017 Norah Ali Al- moneef 94 Free body diagram 22 May 2017 Norah Ali Al- moneef 95 Example : A lift with its load has a mass of 2000 kg. It is supported by a steel cable. Find the tension in he cable when it: a -is at rest b - accelerates upwards uniformly at 1m/s2 C - move upwards at a steady speed of 1 m/s d - moves downwards at a steady speed of 1 m/s e - accelerates downwards with uniform acceleration of 1 m/s2 a) When at rest we can use Newton’s first law which says that the resultant force on the lift is zero. Force acting down is the lifts weight, the force acting up is the tension in the cable. These two must be equal and opposite to give a resultant force of zero. 22 May 2017 Norah Ali Al- moneef 96 b) As the lift is accelerating upwards so T must exceed the weight mg. So the resultant acceleration force by Newton second law, F = ma, so c) As in (a), by Newton’s first law, the resultant force on the lift must be zero, so 22 May 2017 Norah Ali Al- moneef 97 d) As in (c) the tension in the cable will still equal mg since the change in direction of motion does not alter the fact that there is no acceleration. e) If the lift accelerates downwards, then mg must exceed the tension T. So the resultant accelerating force is By Newton’s second law F = ma, so 22 May 2017 Norah Ali Al- moneef 98 example • In Fig. a, a cord holds stationary a block of mass m = 15 kg, on a frictionless plane that is inclined at angle = 27°. (a) What are the magnitudes of the force T on the block from the cord and the normal force N on the block from the plane? T N Fg 0 T 0 mg sin 0 T mg sin (15 kg) (9.8 m / s 2 )(sin 27 ) 67 N N mg cos 0 N mg cos 22 May 2017 (15 kg) (9.8 m / s 2 )(cos 27 ) 131 N Norah Ali Al- moneef 99 Example Find the acceleration of the boxes when the system below is released. Friction is negligible. Same acceleration for both (they move together) 2m m 1. Draw free body diagram for both boxes. 35o 2. Select axes 3. Write Newton’s 2nd law 22 May 2017 Norah Ali Al- moneef 100 N T y 2m y T x m 2mg mg T mg ma 2mg sin T 2ma N 2mg cos 0 35o T mg a 2mg sin m g a 2ma 2g sin g a 2a g g a 2sin 1 3 2sin35 1 0.049 g 3 22 May 2017 Norah Ali Al- moneef 0.48 m/s2 101 a y 2m If θ <30o, a < 0 y x m θ Write Newton’s 2nd law for the whole system 2mg sin mg 3ma 2sin 1 a g 3 y N T y 2m T x m 2mg 35o mg 22 May 2017 Norah Ali Al- moneef 102 Example : a 2.0 kg cookie tin is accelerated at 3.0 m/s2 in the direction shown by a , over a frictionless horizontal surface. • The acceleration is caused by three horizontal forces, only two of which are shown: F 1 of magnitude 10 N and F 2 of magnitude 20 N. What is the third force F 3 in unit-vector notation and as a magnitude and an angle? F1 F2 F3 ma , F3 ma F1 F2 . • Along the x axis we have F3, x ma x F1, x F2, x m (a cos 50 ) F1 cos (150 ) F2 cos 90 22 May 2017 Norah Ali Al- moneef 103 F3, x (2.0 kg) (3.0 m / s 2 ) cos 50 (10 N) cos ( 150 ) (20 N) cos 90 12.5 N F3, y ma y F1, y F2, y m ( a sin 50 ) F1 sin( 150 ) F2 sin 90 (2.0 kg) (3.0 m / s 2 ) sin 50 ( 10 N) sin ( 150 ) (20 N) sin 90 10.4 N F3 F3, x î F3, y ĵ (12.5 N) î (10.4 N) ĵ F3 F3, x F3, y 16 N tan 22 May 2017 2 1 F3, y F3, x 2 40 Norah Ali Al- moneef 104 Example : • Ali pulled a railroad cars (with his teeth) on his end of the rope with a constant force that was 2.5 times his body weight, at an angle of 30° from the horizontal. His mass m was 80 kg. The weight W of the cars was 700 kN, and he moved them 1.0 m along the rails. Assume that the rolling wheels encountered no retarding force from the rails. What was the speed of the cars at the end of the pull? 22 May 2017 Norah Ali Al- moneef 105 SOLUTION: We first find the acceleration of the cars in the x direction : Fnet ,x Ma T cos Ma T 2.5 mg (2.5) (80 kg) (9.8 m / s 2 ) 1960 N The mass M of the cars is :• W 7.0 x 10 5 N 4 M 7 . 143 x 10 kg 2 g 9.8 m / s T cos (1960 N) (cos 30 ) 2 a 0 . 02376 m / s M 7.143 x 10 4 kg v v0 2 a ( x x 0 ) 2 2 v 2 0 2 ( 0.02376 m / s 2 ) (1.0 m) 22 May 2017 Norah Ali Al- moneef v 0.22 m / s 106 22 May 2017 Norah Ali Al- moneef 107 Example: A body of mass 5kg lies on a smooth horizontal table. It is connected by a light inextensible string, which passes over a smooth pulley at the edge of the table, to another body of mass 3kg which is hanging freely. The system is released from rest. Find the tension in the string and the acceleration. Since the string is inextensible and the two bodies are connected, they will both accelerate at the same rate. If the body hanging freely accelerates at a certain rate, the string will pull the other body so that it accelerates at the same rate. The tension in the string on both sides of the pulley will also be the same. This is because the pulley is smooth. Remember, W = mg. The mass of the first body is 5 kg and so its weight is 5g N. The mass of the second body is 3 kg and so its weight is 3g N. Remember to mark in the normal reaction force, a consequence of Newton's Third Law. 22 May 2017 Norah Ali Al- moneef 108 22 May 2017 Norah Ali Al- moneef 109 22 May 2017 Norah Ali Al- moneef 110 22 May 2017 Norah Ali Al- moneef 111 22 May 2017 Norah Ali Al- moneef 112 Example: 22 May 2017 Norah Ali Al- moneef 113 22 May 2017 Norah Ali Al- moneef 114 Example: When two objects with unequal masses are hung vertically over a light, frictionless pulley as in the figure,. Calculate the magnitude of the acceleration of the two objects and the tension in the string. 22 May 2017 Norah Ali Al- moneef 115 Example: 22 May 2017 Norah Ali Al- moneef 116 Solution (a) Both blocks must experience the same acceleration because they are in contact with each other and keep so. Thus we model the system of both blocks as a particle under a net force. For the force is horizontal, we have Fx F ( m1 m2 ) a (1) F So the magnitude of the acceleration reads a . (m1 m2 ) (b) It is obviously the contact force exerted on m2 is in the positive x direction, and the net force acting on the m2 is F2x F2c (2) From Newton's second law, we get m2 F F2x F2c m2 a . (m1 m2 ) 22 May 2017 Norah Ali Al- moneef 117 Example: 22 May 2017 Norah Ali Al- moneef 118 22 May 2017 Norah Ali Al- moneef 119 Example: 22 May 2017 Norah Ali Al- moneef 120 Example: 22 May 2017 Norah Ali Al- moneef 121 Example: 22 May 2017 Norah Ali Al- moneef 122 22 May 2017 Norah Ali Al- moneef 123 Example: Equilibrium and Forces • It is much more difficult for a gymnast to hold his arms out at a 450 angle. • To see why, consider that each arm must still support 350 N vertically to balance the force of gravity. • Use the y-component to find the total force in the gymnast’s left arm. 22 May 2017 Norah Ali Al- moneef 124 Forces in Two Dimensions • The force in the right arm must also be 495 N because it also has a vertical component of 350 N. • When the gymnast’s arms are at an angle, only part of the force from each arm is vertical. • The total force must be larger because the vertical component of force in each arm must still equal half his weight. 22 May 2017 Norah Ali Al- moneef 125 Example: θ = 23.5o Find the component of gravity acting into the plane, and the component acting down along the plane: N = mg cos(θ) = (4.5 kg)(9.81 N/kg) cos(23.5o) = 40.48 N Fx = mg sin(θ) = (4.5 kg)(9.81 N/kg)sin(23.5o) = 17.60 N The plane is frictionless, so the component of gravity parallel to the plane is unopposed. What is the acceleration of the block down the plane? F = ma, -17.60 N = (4.50 kg)a a = F/m = Fx/m = (-17.58 N)/(4.50 kg) = -3.91 m/s2 22 May 2017 Norah Ali Al- moneef 126 Example: + θ = 23.5o What force would make the block accelerate up the plane at 2.10 m/s2? N = 40.48 N, Fx= 17.60 N, F = ma, -17.60 N + F = (4.50 kg)(+2.10 m/s2) F = 27.053 = 27.1 N 22 May 2017 Norah Ali Al- moneef 127 Example: + θ = 23.5o Suppose it accelerates down the plane at 2.71 m/s2. What other force is acting on the block? What is the direction? N = 40.48 N, Fx = 17.60 N, -17.60 N + F = (4.50 kg)(-2.71 m/s2) F = 5.41 N (up the plane) 22 May 2017 Norah Ali Al- moneef F = ma, 128 + θ = 23.5o The block starts from rest and accelerates through a distance of 1.24 m down the plane in .697 s. What other force must be acting along the plane besides gravity? N = 40.48 N, Fx = 17.60 N, F = ma, Δx = v t + 1/2at2 , a = 2 Δx /t2 = -5.1049 m/s2 -17.60 N + F = (4.50 kg)(-5.1049 m/s2) F = -5.3692 N = -5.37 N (down the plane) 22 May 2017 Norah Ali Al- moneef 129 example 22 May 2017 Norah Ali Al- moneef 130 22 May 2017 Norah Ali Al- moneef 131 example 22 May 2017 Norah Ali Al- moneef 132 22 May 2017 Norah Ali Al- moneef 133 Example The maximum load that can safely be supported by a rope in an overhead hoist is 400 N. What is the maximum acceleration that can safely be given to a 25-kilogram object being hoisted vertically upward? Free Body Diagram Frope=400 N W = mg = 25 kg * 9.8 m/s2 Frope a = Fnet/m = (400 N – 245 N)/25 kg a = 6.2 m/s2 W 22 May 2017 Norah Ali Al- moneef 134 Example An object with a mass of 4.0 kg travels with a constant velocity of 4.8 m/s northward. It is then acted on by a force of 6.5 N in the direction of motion and a force of 9.5 N to the south, both of which continue even after the mass comes momentarily to rest. (a) How far will the object travel before coming to rest? (b) What will be its position 2.5 s after the object comes momentarily to rest? M = 4 kg, vo = 4.8 m/s, F2 = 9.5 N south. F1 = 6.5 N north, Solve for x when v = 0: v 2 2 2 o x = vot + ½ at or v = vo + 2ax. Fnet = ma 6.5 N – 9.5 N = 4 kg x a, a = -0.75 m/s2 This means that the acceleration is acting against the initial velocity! 22 May 2017 Norah Ali Al- moneef F1 M F2 135 ’ when v = 0. 0 = (4.8 m/s)2 + 2(-0.75 m/s2)x X = 15 m (b) Where will the object be 2.5 s later. We know its at x =15 with no velocity and -0.75 m/s2 acceleration. x = vot + ½ at2 = 0 m/s x 2.5 s + ½ (-0.75 m/s2)(2.5 s)2 = -2.3 m, So relative to its starting point its at x = 13 m 22 May 2017 Norah Ali Al- moneef 136 Example A 3.50-kg block on a smooth (frictionless) tabletop is attached by a string to a hanging block of mass 2.80 kg, as shown in the Figure. The blocks are released from rest and allowed to move freely. • Find the acceleration of the blocks and the tension in the string. + T T Mg N T=Ma NMg=0 mg Tmg=- ma mg=ma+T=(m+M)a a = mg/(m+M) 22 May 2017 Norah Ali Al- moneef 137 Example I pull a 5 kg mass up with a rope, so that it accelerates 2 m/s2. What is the tension in the rope? T = 59 N 22 May 2017 Norah Ali Al- moneef 138 22 May 2017 Norah Ali Al- moneef 139 22 May 2017 Norah Ali Al- moneef 140 Example Two blocks sit on a frictionless table. The masses are M1=2 kg and M2=3 Kg. A horizontal force F=5 N is applied to Block 1. 1. What is the acceleration of the blocks? 2. What is the force of block 1 on block 2? F M1 M2 1. a = 1 m/s2 2. F21= 3 N 22 May 2017 Norah Ali Al- moneef 141 Example a) Find acceleration b) Find T, the tension in the string 7 kg 5 kg a) a = g/6 = 1.635 m/s2 b) T = 57.2 N 22 May 2017 Norah Ali Al- moneef 142