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Unformatted text preview: Two books laid the foundations for our understanding of forces and motion ... CHAPTER 4 NEWTON’S LAWS • Little bit of history • Forces • Newton’s 1st Law • Newton’s 2nd Law o mass and weight • Newton’s 3rd Law o action reaction pairs o free body diagrams o analysis of day-to-day situations Forces A force is best described as an action or influence that when acting on an object, will change its motion. There are two main types of forces: • contact forces (push, pull, friction, etc.) ⇒ direct force • non-contact (or field) forces (gravity, magnetic, electric, etc.) ⇒ “action” at a distance and long range ... but they are treated in exactly the same way. Dimension [M][L] [ T]2 (show later) Unit: Newton (N) 1N ⇒ weight of a medium size apple, i.e., the downward force exerted on your hand when you hold it. Force is a vector, i.e., with both direction and magnitude and so to find the resultant or the net force when several forces act on a object, they must be added vectorially. r FR r F2 r F2 r F1 r F1 r F2 r F1 r FR r F2 r F1 Question 1: An object on a horizontal table is acted on r r r by three forces, F1 = 4 N, F2 = 3N and F3 = 5N, as shown r F3 below. What is the net force acting on the object? r FR r F3 r F3 r rrr The net force is FR = ∑i Fi = F1 + F2 + ... y r F r Fx = F cos θ r Fy = F sin θ The components of the net force are: FRx = ∑ Fx and FRy = ∑ Fy . 30o r F2 r F θ r F1 x 60o r F1 y r F3 r F2 30 o 60o Example of forces: x Define x and y axes. Then, the net force in the x Two forces: the weight of the ball downward (due to the attraction of the Earth) is balanced by the “normal” force of the ground on the ball. No net force ⇒ no motion r N r W direction is Fx = ∑i Fxi and the net force in the y r F direction is Fy = ∑i Fyi . ∴ Fx = F1 cos 30o + F2 cos 60o − F3 = 3.64 + 1.50 − 5.00 = 0.14 N, and Fx ˆ i θ Fy ˆ j r FR r W r FR Fy = F1 sin 30o + F2 sin 60o Two forces: the weight of the ball downward and the force of the foot on the ball. Net resultant force ⇒ change in motion acceleration = 2.00 − 2.60 = −0.60 N. Thus, the net force is r F = (0.14ˆ − 0.60ˆ ) N, i j R with magnitude r FR = (Fx )2 + (Fy )2 = (0.14)2 + ( −0.60)2 = 0.62 N . The angle θ = tan −1 Fy = tan −1 ( −4.29) = −76.9o . Fx r W One force: the weight of the ball downward. Net force ⇒ change in motion acceleration Important concept ... ... we cannot neglect the effect of the Earth ... it always produces a non-contact force! Analysis of moving objects led to ... A “simple” example .... Newton’s 1st Law: 2 A body at rest, or moving with constant velocity, will 1 remain at rest, or moving with constant velocity, unless it is acted on by a net force. (NB: the reference frame is 3 the Earth.) Sometimes called the Law of inertia ... 1 The force on you due to the Earth, i.e., your weight, • Tendancy for a body that’s moving to keep moving. (downward) is balanced by the force of the seat of the • Tendancy for a body at rest to remain at rest. chair on you (upward). No net force. r ∴ If the sum of forces ∑i F , i.e., the net force, on an object is zero it remains at rest or continues moving with constant velocity. ( ∑ Fx = 0 and ∑ Fy = 0.) 2 The force on the ice cream cone due to the Earth (downward) is balanced by the force of your hand on the cone (upward). No net force. This is called EQUILIBRIUM. 3 The total weight of you, the ice cream cone and the Implication ⇒ to change motion we have to apply a force ... Newton’s 2nd Law gives us the relationship between the force and the change in motion. chair (downward) is balanced by the force of the ground on the chair legs (upward). No net force. But don’t get fooled!! some common misconceptions .... Are you “thrown” backwards when a car accelerates from rest? NO!! You tend to remain at rest as the seat moves forward! DISCUSSION PROBLEM [4.1]: Why does a wire stretched between two posts always sag in the middle no matter how tightly it is stretched? r F Are you “thrown” outwards when a car turns a corner? NO!! As the car turns, you tend to continue to move in a straight line; so the side of the car exerts a force on you! Newton’s 2nd Law: So, to produce a change in motion, there must be a net force ... In the Principia Newton stated that The change in motion is proportional to the motive r F force impressed; and is made in the direction of r v the right line in which that force is impressed. r F r v ... to move an object from rest, i.e., to change its velocity, r v r F frictional force opposing motion or slow down a puck or change the direction of a ball. r F r v1 Today, we write the 2nd Law as r r dp F= , dt r where p = mv , which today is called the momentum; Newton called it the quantity of motion. r r r r dp d ( mv ) dv r dm r r dm ∴F = = =m +v = ma + v . dt dt dt dt dt However, most textbooks ignore the full Newtonian expression and assume that m is constant. So, you will find Newton’s 2nd Law usually stated this way: r v2 If a net force acts on an object it accelerates (i.e., there is a change in motion). The direction of the acceleration is the same as the net force, r r i.e., ∑i Fi = ma . And in component form, The constant, m, in the expression ∑ Fx = ma x : ∑ Fy = ma y : ∑ Fz = ma z F = ma where m is a constant. Remember, however, these is really a measure of how difficult it is to change the expressions are only true if the mass remains constant. motion of an object. We call it the inertial mass, or (This form of the Law would not apply therefore to a simply, the mass. Note: mass is a scalar quantity. rocket that is losing mass by burning fuel.) To understand that concept, think about how you check Note: the “weight” of something ... • The net force has to be an external force, i.e., not inside the object or system, e.g., you can’t lift yourself! r • The product ma is not itself a force ... it is the net r force that is put equal to ma . The dimension of force is the same as [L] [M][L] mass × acceleration ⇒ [M] 2 = . [ T] [ T]2 The unit of force is the Newton (N). m2 m1 ... you check to see “how easy or difficult it is to make them move”! For the same acceleration ... F F a = 1m = 2m , 1 2 i.e., m1 F1 m2 = F2 So, in fact, you are comparing the masses ! OK ... so, a force changes motion. But when you throw a baseball, what keeps it moving to the right after it leaves your hand? r w Question 2: An object of mass m is traveling at an initial speed v o = 25.0 m/s. A net force of 15.0N acting on the object in a direction opposing the motion, causes it to The answer is ... inertia ... remember Newton’s 1st Law. Neglecting air resistance, there are no horizontal forces to change its horizontal motion, so it keeps going to the right! Note, inertia is NOT a force. But, there is a force downward (its weight), which causes the ball to “fall”. These effects combine so the ball follows a parabolic trajectory. come to a stop in a distance of 62.5m. What is the mass of the object? F = 15.0 N + ve direction x v=0 xo v o = 25.0 m/s ( x − xo ) = 62.5 m Define the positive direction to the right. First, we need What’s the connection between mass and weight ? • MASS: inertial property larger mass ⇒ larger force required to accelerate. to find the acceleration. From earlier: ∴a = v 2 = v o2 + 2a ( x − x o ) v 2 − v o2 (25.0)2 =− = −5.00 m/s2 2( x − x o ) 2 × 62.5 Now, using Newton’s 2nd Law F = ma , note that F opposes the motion, i.e., it is in the − ve direction. F −15.0 ∴m = = = 3.00 kg. a −5.00 • WEIGHT: the force exerted by the Earth on an object. Example: Bowling ball ⇒ difficult to lift (weight) ⇒ difficult to throw (mass) Why does an object fall when you let go of it? Don’t say ... “because of gravity” ... that doesn’t ANSWER the question We have two forces acting on the ball: Force of the hand (up) and the weight of ball (down). Initially, they are equal (but opposite) so there is no net force r i j Question 3: A force F = 6ˆ − 3ˆ N acts on an object of acting on the ball and so it is at rest. mass 1.5kg, initially at rest at the origin. (a) Find the ( ) acceleration vector of the object. (b) What is the When the ball is released, the hand applies magnitude of the accceleration? (c) What is the velocity no force but the weight still exists, so there is a of the object after 2.0s? (d) What is the displacement now a net force downward . Therefore, after 2.0s? according to Newton’s 2nd Law, the object must change its motion ... it will accelerate downward. Using Newton’s second Law the downward force is r r F = ma , rr with a = g (acceleration due to gravity). r This force ( = mg ) is the weight. ˆ j ˆ i r F = (6ˆ − 3ˆ ) N i j 1.5kg r r (a) Newton’s 2nd Law: F = ma r i j r F (6ˆ − 3ˆ ) ∴a = = = (4ˆ − 2ˆ ) m/s2. i j m 1.5 Note that the acceleration is in the same direction as the rr force (i.e., a F ). r (b) a = a x2 + a y2 = 42 + ( −2)2 = 20 = 4.47 m/s2. rrr i j i j (c) v = v o + at = 2(4ˆ − 2ˆ ) = (8ˆ − 4ˆ ) m/s . r v = 82 + ( −4)2 = 8.94 m/s 1r 1 rr r i j i j (d) ( r − ro ) = v o t + at 2 = (4ˆ − 2ˆ )(2)2 = (8ˆ − 4ˆ ) m. 2 2 rr ( r − ro ) = 82 + ( −4)2 = 8.94 m. If you analyze a situation involving forces you’ll find that forces ALWAYS occur in pairs: r FBT r FFB r FTB r FBF (Spring) Books on table (down) Table on books (up) r FAB r FBA A B r Pulling force A (to the left) is applied on B ( Fr ). AB Pulling force B (to the right) is applied on A ( FBA ). r FAW r FWA A r Pulling force A (to the left) is applied on the wall ( FAW ). r The wall applies a force (to the right) on A ( FWA). Who pulls harder, i.e., who applies the greater force, in the scenario shown here? r FAB r FBA Newton’s third Law: A B If an object A exerts a force on an object B (an action), object B exerts an equal and opposite force on A (the FAB FBA 50 FBA 0 −50 reaction), i.e., r r FAB = −FBA r r ( FAB = FBA ) Note: FAB 0 2 46 Time (s) 8 10 • the two forces act on different objects, sometimes called an action/reaction pair. Ha! They both “pull” equal amounts ... the force exerted by A on B is equal and opposite to the force exerted by • the Law applies whether the objects are at rest, B on A, i.e., r r FAB = −FBA and FAB = FBA at all times! moving with constant velocity or accelerating. This leads us to ... Often, Newton’s laws appear in combination look ... r FBT r FTB Books on table (down) Table on books (up) r r FBT = FTB r FFB r FFE r FBF r r FFB = FBF Centripetal force (Action) Car on you Reaction (“Centrifugal force”) You on car r FEF r r FFE = FEF That’s weird! How come it seems to hurt your eye more than it hurts your finger? But we have to be careful in identifying action/reaction pairs correctly (later). First Law: “Every body perseveres in its state of rest, or of uniform motion in a right [straight] line, unless it is compelled to change that state by forces impressed thereon.” Third Law: “To every action there is always opposed an equal reaction: or the mutual action of two bodies on each other are always equal, and directed to contrary parts.” Principia Mathematica (1687). So, if the Earth applies a force on an apple (its weight), the apple must apply an equal and opposite force on the Earth! Does that mean the Earth moves towards an apple when r FEA r FAE Take something as simple as a book or cups on a table on the Earth (i.e., the floor) ... ~ Let’s identify ALL action-reaction pairs ~ we drop the apple? Let’s see ... r r We know FEA = FAE , and the 2nd Law tells us: FEA = mAa A and FAE = mE a E . ma ∴ mA a A = mE a E ⇒ a E = A A . mE But mE ≈ 6 × 1024 kg , mA ≈ 0.1 kg and a A = g ∴ a E ≈ 1.7 × 10 −25 m/s2. r r r • Forces acting ON the book: FTB and FEB ( = w B ) These are not an action-reaction pair Why ?? r r r • Forces acting ON the table: FBT and FET ( = w T ) If the apple takes 1s to reach Earth ... 1 ∆y A = a A t 2 ≈ 4.91m. 2 1 ∆y E = a E t 2 ≈ 0.85 × 10 −25 m. 2 Boy ... that’s small! r r • Forces acting ON the Earth: FBE and FTE For action-reaction pairs: r r F12 = F21 . Does Newton’s 3rd Law apply if objects are moving? You betcha!! You know all about the birds and bees, but now it’s time to learn about FBD’s Believe me, dear, crucial to solving “force problems” is the free-body diagram. W R S Identify all the forces associated with the rope: (Take L ⇒ R as positive direction) r FRS (rope on sled)* r FSR (sled on rope)* r FWR (woman on rope)# r FRW (rope on woman)# r r r r But FRS = −FSR and FWR = −FRW . r r The net force on the rope ⇒ FWR − FSR . r r So, if FWR > FSR + A free-body diagram (FBD) shows all of the forces − acting ON an object. For example, the forces acting on a + − the rope and everything attached to it, moves to the right, but the equations above still hold! ball that’s been kicked ... r F r N r W ⇒ r W r W ⇒ ⇒ STRATEGY for analyzing “force” problems: Draw free body diagram (identify the forces acting ON an object) Newton’s third Law Question 4: A ball, weighing 100N, is suspended using a system of strings, as shown below. What is the tension in the three strings? 45o Equilibrium Motion (i.e., at rest or constant velocity) Newton’s first Law 90o (i.e., accelerating) Newton’s second Law Separate the force components in two (perpendicular) directions. 100N An alternative but equivalent solution: 45o 90 o Since the knot is T3 45o T3 T2 T2 T1 45o the point through which all the forces T1 act, we shall 100N 90o r T3 r T2 r r T1 = mg consider the free-body diagram at the position of the knot. 100N r T3 45o B r T1 r A T2 C r r r T1 + T2 + T3 = 0 rr r Identify the forces acting on the knot, T1, T2 and T3. Use the horizontal and vertical components of the forces. Since the knot is not moving, ∑ Fh = 0 and ∑ Fv = 0 (h) T2 = T3 cos 45o . T ∴ T2 = 3 (v) T1 = T3 sin 45o . T3 ∴ T1 = 2 2 . Since the knot is stationary there is no net force acting r r r r on the knot, i.e., Fnet = T1 + T2 + T3 = 0. r r r ∴ − T1 = T2 + T3. Therefore, the force diagram is closed, right-angled triangle, shown on the right. Since ∠ACB = 90o and . But T1 ⇒ weight of ball = 100N. Also, T1 = T2. ∴ T2 = 100 N and T3 = 2 × T1 = 141N. ∠ABC = ∠BAC = 45o , the triangle is isosceles, so AC = BC and AB = 2 × BC. r r ∴ T1 = T2 = 100 N, r and T3 = 2 × 100 N = 141N. N (Upward force due to spring balance, i.e., apparent weight. Question 5: A 2kg mass hangs from a spring scale, calibrated in Newtons, that hangs from the ceiling of an Draw the free body w = mg (Force due to Earth) block. diagram for the 2kg when the elevator is (a) moving up In (a) and (b) the block is not accelerating. ∴ a v = 0 , r so ∑ Fv = 0. with a constant velocity of 30m/s, Take up direction as positive: elevator. What does the scale read (b) moving down with a constant velocity of 30m/s, and (c) is ∴ N +( − mg ) = 0 , i.e., N = mg = 19.8 N ... (2 kg). ascending at 20m/s and gaining r (c) Now we have: ∑ Fv = ma v . speed speed at the rate of 3m/s2? Then N +( − mg ) = ma v ... a v > 0 (d) From t = 0 to t = 5s, the i.e., N = m( g + a v ) = 25.6 N ... (2.61 kg). elevator moves up at 10m/s. Its velocity is then reduced uniformly to zero in the next 4s, so that it comes to rest at t = 9s. What does the scale read during the time interval 0 < t < 9s? (d) For 0 < t ≤ 5s, a v = 0. ∴ N = mg = 19.8 N. For 5 < t ≤ 9s, a v = ∆v ∆t = ( 0 − 10 ) 4 = −2.5 m/s2 . Since N +( − mg ) = ma v ... (now a v < 0) ∴ N = m( g + a v ) = 14.6 N ... (1.49 kg) This is ME ... ! Draw the free body diagram for the forces on me: N (Upward force from scale) DISCUSSION PROBLEM [4.2]: w = mg (force due to Earth) ∑ v Fv = N − mg = ma v . My apparent weight is the force I exert on the scale What is the apparent weight of a person in an elevator if the cable breaks? ⇒ N = m( g + a v ) . 4.0 a v (m/s ) 2.0 2 acceleration upward Yikes! 0 acceleration downward − 2.0 − 4.0 80 120 160 200 240 Apparent weight (pounds) So, if you know your true weight, you can find the acceleration of an elevator! y y x N T T x m2 g m1g Treat each object separately. Note that T, the tension, is common to both free-body diagrams. Assume m2 descends (i.e., m1 ascends the slope). Question 6: Two objects are connected by a massless [#1] along x: T + ( − m1g sin θ ) = m1a string, as shown above. The incline and pulleys are [#2] along y: T + ( − m2 g ) = − m2a frictionless. Find the acceleration of the objects and the tension in the string for (a) general values of m1, m2 and θ , and (b) for m1 = m2 = 5kg and θ = 30o . ... (i) ... ... (ii) Subtract (ii) from (i): ( m2 − m1 sin θ ) g = ( m1 + m2 )a ∴a = ( m2 − m1 sin θ ) g . ( m1 + m2 ) Note: if m2 > m1 sin θ then a > 0, if m2 < m1 sin θ then a < 0, if m2 = m1 sin θ then a = 0. Substitute a= ( m2 − m1 sin θ ) g ( m1 + m2 ) in equation (ii). Then T − m1g sin θ = ∴T = m1( m2 − m1 sin θ)g . ( m1 + m2 ) m1( m2 − m1 sin θ)g + ( m1 + m2 )m1g sin θ , ( m1 + m2 ) i.e., T = m1m2 (1 + sin θ ) g . m1 + m2 Take m1 = m2 = 5kg and θ = 30o . ∴a = ( 5 − 5 × 0.5) × 9.81 = 2.45 m/s2. ( 5 + 5) Since a > 0, m1 moves up the slope. Calculate the tension, T, from above ... T= 5 × 5 × (1 + sin 30o ) × 9.81 = 36.8 N, 5+5 or from equation (ii). Question 7: Two blocks are in contact on a frictionless, r horizontal surface. A horizontal force F is applied to one of them, as shown. Find the acceleration and the contact force for (a) general values of F, m1 and m2, and (b) for F = 3.2 N, m1 = 2 kg and m2 = 6 kg. N1 N2 F F21 m1g F12 m2 g Draw the free-body diagrams for each object. (a) Consider only horizontal forces (why??) ... #1 ... F − F21 = m1a . #2 ... F12 = m2a . But F12 = F21. ∴a = F . m1 + m2 The contact force is: F12 ( = F21 ) = (b) a = m2 F. m1 + m2 3.2 F = = 0.400 m/s2 . m1 + m2 2 + 6 F12 = 6 × 3.2 = 2.4 N. 2+6 ...
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