* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download CHAPTER 4 NEWTON`S LAWS • Little bit of history • Forces
Equations of motion wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Coriolis force wikipedia , lookup
Classical mechanics wikipedia , lookup
Nuclear force wikipedia , lookup
Fundamental interaction wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Centrifugal force wikipedia , lookup
Classical central-force problem wikipedia , lookup
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 ! mass and weight • Newton’s 3rd Law ! action reaction pairs ! free body diagrams ! the normal force ! 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. ! FR ! F2 ! F2 ! F1 ! F1 ! F2 ! F1 ! FR Question 4.1: An object on a horizontal table is acted ! ! ! on by three forces, F1 = 3 N, F2 = 4 N and F3 = 3 N, as ! F2 ! F1 ! F3 shown below. What is the net force acting on the object? ! FR ! F3 ! F3 ! ! ! ! The net force is FR = ∑i Fi = F1 + F2 + ... y ! F ! Fx = F cos θ ! Fy = F sin θ The components of the net force are: FRx = ∑ Fx and FRy = ∑ Fy . 30" 60" ! F2 ! F θ ! F1 x Example of forces: y " F3 = 3 N Define x and y axes. Then, " F1 = 3 N the net force in the x 30! x 60! " F2 = 4 N direction is Fx = ∑i Fxi and the net force in the y ! W direction is Fy = ∑i Fyi . ! ! F ! ∴Fx = F1 cos30 + F2 cos60 − F3 = 2.60 N + 2.00 N − 3.00 N = 1.60 N, and 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 ! N Fy = F1 sin 30! + F2 sin 60! ! W ! FR = 1.50 N − 3.46 N = −1.96 N. Fx ˆi θ Fy ˆj " FR Two forces: the weight of the ball downward and the force of the foot on the ball. Net resultant force ⇒ change in motion acceleration Thus, the net force is " F = (1.60 ˆi − 1.96 ˆj) N, R with magnitude " FR = (Fx )2 + (Fy )2 ! W = (1.60 N)2 + (−1.96 N)2 = 2.53 N. ⎛F ⎞ −1 ! The angle θ = tan −1 ⎜ y ⎟ = tan (1.225) = 50.8 . F x⎠ ⎝ 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: A body at rest, or moving with constant velocity, will 2 1 remain at rest, or moving with constant velocity, unless it is acted on by a net force. (NB: the reference frame is the Earth.) Sometimes called the Law of inertia ... 3 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. ! ∴ 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.) This is called EQUILIBRIUM. Implication ⇒ to change motion we have to apply a net force ... Newton’s 2nd Law gives us the relationship between the force and the change in motion. 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. 3 The total weight of you, the ice cream cone and the 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!! There is no “throwing” force; 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 Are you “thrown” ! F outwards when a car turns a corner? NO!! There is no “throwing” force; as the car turns, you tend to continue to move in a straight line; so the side of the car exerts an inwards force on you! in the middle no matter how tightly it is stretched? So, to produce a change in motion, there must be a net force ... Newton’s 2nd Law: In the Principia Newton stated that The change in motion is proportional to the motive force impressed; and is made in the direction of ! F the right line in which that force is impressed. ! v ! F ! v ... to move an object from rest, i.e., to change its velocity, ! v ! F frictional force opposing motion or slow down a puck or change the direction of a ball. ! F ! v1 Today, we write the 2nd Law as ! dp! F= , dt ! where p = mv , which today is called the momentum; Newton called it the quantity of motion. ! ! dp! d(mv! ) dv ! dm ! ! 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: ! 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, ! ! i.e., ∑ i Fi = ma . And in component form, ∑ Fx = ma x : ∑ Fy = ma y : ∑ Fz = ma z where m is a constant. Remember, however, these expressions are only true if the mass remains constant. So, this form of the Law does not apply to a rocket that is losing mass by burning fuel. " F3 NOTE: ! • The product ma is not itself a force ... it is the net ! force that is put equal to ma . ! ! • The acceleration a is parallel to the net force F. ˆj 30! 60! ˆi " F2 • The net force has to be an external force, i.e., not inside the object or system, e.g., you can’t lift yourself! " F1 Question 4.2: An object on a horizontal table is acted on by three forces, F1 = 3 N, F2 = 4 N and F3 = 3 N, as shown above. If the mass of the object is 2.0 kg, and it is initially at rest at the origin, (a) what is its acceleration? (b) What is its velocity after 4.0 s? (c) What is its displacement after 4.0 s? 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). y " F3 = 3 N " F1 = 3 N 30! x 60! " F2 = 4 N (a) From before (Q4.1) we found the resultant force was: " FR = (1.60 ˆi − 1.96 ˆj) N. " " Using Newton’s 2nd Law FR = ma . " " FR (1.60ˆi − 1.96ˆj) N ∴a = = m 2.0 kg = (0.80 ˆi − 0.98 ˆj) m/s 2 , and the magnitude of the acceleration is " a = (0.80 m/s2 )2 + (−0.98 m/s2 )2 = 1.27 m/s2 " and it is parallel to FR. " " " (b) v = v ! + at = (0.80 ˆi − 0.98 ˆj) m/s 2 × 4.0 s = (3.20 ˆi − 3.92 ˆj) m/s. " ∴ v = (3.20 m/s) 2 + (−3.92 m/s)2 = 5.06 m/s. 1! 1 ! ! ! (c) r − r" = v " t + at 2 = (0.80 ˆi − 0.98 ˆj) m/s2 × (4.0 s)2 2 2 = (6.40 ˆi − 7.84 ˆj) m. ! ! ∴ r − r" = (6.40 m) 2 + (−7.84 m)2 = 10.1 m. The constant, m, in the expression F = ma is really a measure of how difficult it is to change the motion of an object. We call it the inertial mass, or simply, the mass. Note: mass is a scalar quantity. To understand that concept, think about how you check the “weight” of something ... DISCUSSION PROBLEM [4.2]: We have just seen that the way we choose between the weights of two objects we move them up-and-down and determine how difficult it is to move them. m2 m1 m1 m2 ... 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 ! Will this idea work in a weightless environment? OK ... so, a force changes motion. But when you throw a baseball, what keeps it moving to the right after it leaves your hand? ! w Question 4.3: An object of mass m is traveling at an initial speed v ! = 25.0 m/s. A net force frictional force of 15.0 N acting on the object in a direction opposing the motion, causes it to come to a stop after a distance of The answer is ... inertia ... remember Newton’s 1st Law. 62.5 m. (a) What is the mass of the object? (b) How Neglecting air resistance, there are no horizontal forces long does it take to stop? to change its horizontal motion, so it keeps going to the right! Note, inertia is NOT a force. There is a force downward (its weight), which causes the ball to “fall”. These effects combine so the ball follows a parabolic trajectory. v ! = 25.0 m/s v=0 x! x F = 15.0 N + ve direction (x − x! ) = 62.5 m Define the positive direction to the right. First, we need to find the (negative) acceleration. 2 2 (a) We have: v = v ! + 2a(x − x ! ) ∴a = v 2 − v !2 (25.0 m)2 =− = −5.00 m/s 2 . 2(x − x ! ) 2 × 62.5 m Question 4.4: Two forces, F1 and F2, act on an object on a frictionless surface, as shown. In which case is the acceleration greatest? F2 = 3 N Note that F opposes the motion, i.e., it acts in the −ve i.e., m = F −15.0 N = = 3.00 kg. a −5.00 m/s2 (b) We have v = v ! + at, v − v ! 0 − 25.0 m i.e., t = = = 5.00 s. a −5.00 m/s2 F1 = 12 N F2 = 4 N A direction (same as the acceleration). Using Newton’s 2nd Law, F = ma , 6 kg F2 = 2 N 3 kg C 4 kg F1 = 8 N B F1 = 8 N F2 = 2 N 2 kg D F1 = 8 N F2 = 3 N 6 kg F1 = 12 N F2 = 4 N A F2 = 2 N 3 kg F1 = 8 N 4 kg B F1 = 8 N F2 = 2 N 2 kg C F1 = 8 N D The net force on a block in each case is What’s the difference between mass and weight ? • MASS: inertial property larger mass ⇒ larger force required to accelerate. F1 − F2 = F. By Newton’s 2nd law, that force produces an acceleration given by F = ma , i.e., a = F m. 9N 4N ∴a A = = 1.5 m/s2 : a B = = 1 m/s2 6 kg 4 kg aC = 6N 6N = 2 m/s2 : a D = = 3 m/s2 3 kg 2 kg So, the answer is D. • WEIGHT: the force exerted by the Earth on an object. Example: Bowling ball ⇒ difficult to throw (mass) ⇒ difficult to lift (weight) If you analyze a situation involving forces you’ll find that forces ALWAYS occur in pairs: ! FBT Who pulls harder, i.e., who applies the greater force, in the scenario shown here? ! FAB ! FFB ! FTB ! FBA A ! FBF (Spring) Books on table (down) Table on books (up) ! FAB ! FBA A B ! Pulling force A (to the left) is applied on B ( FAB ). ! Pulling force B (to the right) is applied on A ( FBA ). ! FAW ! FWA A ! Pulling force A (to the left) is applied on the wall ( FAW ). ! The wall applies a force (to the right) on A ( FWA ). B FAB FBA 50 FBA 0 −50 FAB 0 2 4 6 Time (s) 8 10 Ha! They both “pull” equal amounts ... the force exerted by A on B is equal and opposite to the force exerted by B on A, i.e., ! ! FAB = − FBA and FAB = FBA at all times! This leads us to ... Examples of action/reaction pairs: ! FBT ! FTB Newton’s third Law: If an object A exerts a force on an object B (an action), Books on table (down) Table ! on books ! (up) FBT = FTB object B exerts an equal and opposite force on A (the reaction), i.e., ! ! FAB = − FBA ! ! ( FAB = FBA ) Note: • the two forces act on different objects, sometimes ! FFB ! FFE ! FBF ! ! FFB = FBF ! FEF ! ! FFE = FEF called an action/reaction pair. • the Law applies whether the objects are at rest, moving with constant velocity or accelerating. 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). Often, Newton’s laws appear in combination look ... 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 Centripetal force (Action) Car on you Reaction (“Centrifugal force”) You on car we drop the apple? Let’s see ... ! ! We know FEA = FAE , and the 2nd Law tells us: FEA = m Aa A and FAE = mE a E . m a ∴mAa A = mE a E ⇒ a E = A A . mE 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). But mE ≈ 6 × 1024 kg, mA ≈ 0.1 kg and a A = g ∴a E ≈ 1.7 × 10 −25 m/s2 . If the apple takes 1 s to reach Earth ... 1 Δy A = a A t 2 ≈ 4.91 m. 2 1 Δy E = a E t 2 ≈ 0.85 ×10 −25 m. 2 WOW ... that’s small! ! FEA ! 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 ~ DISCUSSION PROBLEM [4.3]: You are holding a baseball in your hand. The reaction force to the weight of the baseball is .... ! ! ! • Forces acting ON the book: FTB and FEB (= w B ) These are not an action-reaction pair Why ?? ! ! ! • Forces acting ON the table: FBT and FET (= w T ) ! ! • Forces acting ON the Earth: FBE and FTE For action-reaction pairs: ! ! F12 = F21 . A: the gravitational force of the Earth on the ball, B: the gravitational force of the ball on the Earth, C: the force of your hand on the ball, D: the force of the ball on your hand, or E: the gravitational force of the Earth on your hand. 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) ! FRS (rope on sled)* ! FSR (sled on rope)* ! FWR (woman on rope)# ! FRW (rope on woman)# ! ! ! ! But FRS = − FSR and FWR = −FRW . ! ! The net force on the rope ⇒ FWR − FSR . ! ! 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 ball + − the rope and everything attached to it, moves to the right, but the equations above still hold! that’s been kicked ... ! F ! N ! W ⇒ ! W ! W ⇒ ⇒ The Normal Force In mechanics, the normal force FN (sometimes n or N) is the component of the contact force exerted on an object , DISCUSSION PROBLEM [4.4]: perpendicular to the surface of contact by, for example, Which of the free-body diagrams below best represents the surface of a floor or incline, preventing the object the forces acting on an object sliding down a frictionless from penetrating the surface. In the case of an object FN incline? resting on a horizontal surface (e.g., a table top, the ground), ! v the normal force is equal and mg opposite to the weight of the object (Newton’s 3rd Law), i.e., FN = mg . When an object rests on an incline, the normal force is perpendicular to the plane the FN object rests on. In that case, the A B C D mgcos θ mg θ mgsin θ normal force is equal and opposite to the component of the weight perpendicular to the surface, i.e., FN = mg cosθ. There is a force acting parallel to the incline, i.e., the component of the weight parallel to the surface, mg sin θ. FN mg cos θ mg STRATEGY for analyzing “force” problems: FN mgsin θ θ Draw free body diagram mgcos θ FR mg (identify the forces acting ON an object) Newton’s third Law mgsin θ θ Therefore, when an object sits on a frictionless incline, the resultant (net) force, FR = mg sin θ produces motion down the incline. By Newton’s 2nd Law the acceleration of the object down the slope is given by F a = R = gsin θ, m a result I showed in Chapter 2. Equilibrium Motion (i.e., at rest or constant velocity) Newton’s first Law (i.e., accelerating) Newton’s second Law Separate the force components in two (perpendicular) directions. F2 F1 f Draw the FBD for the blocks. Since the blocks are not Question 4.5: Two forces, F1 and F2, act on an object on accelerating, Newton’s 2nd Law tells us the net force on a surface, as shown. If the objects are all moving with a each block is zero, i.e., constant velocity, in which case is the frictional force F1 − F2 − f = 0 greatest? ∴f = F1 − F2 . ( F1 − F2 ) is largest for block A ( = 9 N to the right). Since F2 = 3 N 6 kg F1 = 12 N F2 = 4 N A F2 = 2 N 3 kg C 4 kg F1 = 8 N the net force must be zero, the frictional force of the ground on block A must be 9 N (to the left). B F1 = 8 N F2 = 2 N 2 kg F1 = 8 N F2 = 3 N f 6 kg F1 = 12 N F2 = 4 N A 4 kg F1 = 8 N B D F2 = 2 N 3 kg C F1 = 8 N F2 = 2 N 2 kg D F1 = 8 N DISCUSSION PROBLEM [4.5]: A Question 4.6: To prevent a box from sliding down a frictionless inclined plane, physics student Anna pushes on the box horizontally with just enough force so that the box is stationary. If the mass of the box is 2.00 kg and the slope of the incline is 35!, what is the magnitude of B the force she exerts? You and a friend have identical cars. You want to rip an old pair of Levi’s apart; there are two possible Anna arrangements, as shown above. Which is more likely to do the job, i.e., in which is the tension in the jeans 35! greater? A: A B: B C: They’re the same Draw the free body diagram for the box: FN y FA x mgcos θ mgsin θ θ Question 4.7: The system shown below is in equilibrium. mg Find the three tensions T1, T2 and T3 in the string, and the The components of the weight force along x and along y unknown mass, M. are (−mg sin θ) and (−mg cosθ), respectively. The components of the force Anna applies (FA ) along x and y are FA cosθ and −FA sin θ, FA sin θ respectively. Summing the FA θ FA cosθ forces in the x-direction, we have θ ∑ Fx = 0 = FA cosθ − mg sin θ, i.e., FA = mg tan θ. ( ) ∴FA = (2.00 kg ) 9.81 m/s 2 (0.700) = 13.7 N. 60! T3 T1 M T2 6 kg 60! First draw the free body diagram for the 6 kg T2 mg mass; the two forces are the weight (mg) and T2. Since the system is in equilibrium, T2 = mg = 6.00 kg × 9.81 m/s2 = 58.9 N. 60! T3 T1 M 60! 60! T3 T1 T2 M The tensions all pass through a single point (circled); draw the free body diagram at that point. Then, we have T2 ∑ Fh = 0 and ∑ Fv = 0. ! −T1 sin 30 + T3 sin 30 = 0. ∴T1 = T3 (v) T1 cos30! + T3 cos30! − T2 = 0, T2 58.9 N ! = 0.866 = 68.0 N. cos30 ∴T1 = T3 = 34.0 N. Since the pulley only changes the direction of T1, it does not affect its magnitude. ! (h) i.e., T1 + T3 = T2 6 kg 6 kg T1 30! 30! T3 60! i.e., ∴T1 = Mg, T 34.0 N M= 1 = = 3.47 kg. g 9.81 m/s2 Sometimes, we may not know the angles as in this problem ... Question 4.8: Isobel Newton hangs motionless by one Identify all of the forces acting through Isobel’s hand and draw the vector force diagram. ! F2 hand from a clothes line, as shown. If the clothes line is ! F1 ! w ! F2 ! w ! F1 about to break, which side is most likely to break? Since she is not moving, the net force on her hand is zero, ! ! ! ! i.e., Fnet = F1 + F2 + w = 0. So the vector diagram shown on the right must be closed. ! ! Clearly, F1 > F2 and so the part of the line on the right has the greater tension and so is more likely to break! A: The left side. B: The right side. C: Equal chance of either side breaking. Hence, even if the angles are not known, you can use the fact that the vector diagram of forces must be a closed figure to solve a problem. N (Upward force due to spring balance, i.e., apparent weight. Question 4.9: A 2 kg mass hangs from a spring scale, calibrated in Newtons, that hangs from the ceiling of an elevator. What does the scale read when the elevator is (a) moving up with a constant velocity of 30 m/s, (b) moving down with a constant velocity of 30 m/s, and (c) is ascending at 20 m/s and gaining Draw the free body diagram for the 2 kg block. In parts (a) av w = mg (Force due to Earth) and (b) the block is not accelerating. ! ∴a v = 0 so ∑ Fv = 0. Take up direction as positive: ∴ N +(−mg ) = 0, i.e., N = mg = (2 kg)(9.81 m/s 2 ) = 19.8 N ... (2 kg). speed at the rate of 3 m/s 2 ? elevator moves up at 10 m/s. Its velocity is then reduced (c) Since the block is accelerating upward we have: ! ∑ Fv = ma v ( a v > 0). uniformly to zero in the next 4 s, so that it comes to rest at Then N +(−mg ) = ma v (d) From t = 0 to t = 5 s, the t = 9 s. What does the scale read during the time interval 0 < t < 9 s? i.e., N = m(g + a v ) = (2 kg)(12.81 m/s 2 ) = 25.6 N ... (2.61 kg). (d) When 0 < t ≤ 5 s, a v = 0, so N = 19.8 N. This is ME ... ! Draw the free body diagram for the forces on me: N (Upward force from scale) w = mg (force due to Earth) When 5 s < t ≤ 9s, a v = Δv (0 − 10 m/s) = = −2.5 m/s2 . Δt 4s Since N +(−mg ) = ma v ∑ v Fv = N −mg = ma v . My apparent weight is the force I exert on the scale ⇒ N = m(g + a v ). ... but a v < 0. 2 ∴ N = m(g + a v ) = (2 kg)(7.31 m/s ) = 14.6 N ... (1.49 kg) 4.0 av (m/s2 ) acceleration upward 2.0 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! Note that the apparent weight of someone in an elevator if the cable breaks is zero. Draw the FBD for the forces acting on the person ... N (Upward force from scale) w = mg (force due to Earth) Yikes! From the free-body diagram N −mg = ma v , so apparent weight N = m( g + a v ) = m( g + (−g)) =0 Conceptually what’s happening is ... the floor and weighing scales are falling at the same rate as the person inside the car so there’s no net force on the weighing scale. This is what is popularly known as weightlessness except, of course, the person still has real weight! Question 4.10: Two objects are connected by a massless string, as shown above. The incline and pulleys are frictionless. (a) Show that the acceleration of the objects is a= (m2 − m1 sin θ)g . (m1 + m2 ) (b) Find the acceleration and tension in the string when m1 = m2 = 5 kg and θ = 30! . y y x N T T y x y m2 g m1g x T N m1g (a) Treat each object separately. Note that T, the tension, is common to both free-body diagrams as the pulley only changes the direction of T. Assume m2 (b) With m1 = m2 = 5 kg and θ = 30! , a= descends (i.e., m1 ascends the slope). [#1] along x: T + (−m1gsin θ) = m1a [#2] along y: T + (−m2 g) = −m2a ... (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. = ( m2 − m1 sin θ)g ( m1 + m2 ) (5 kg − (5 kg × 0.50))(9.81 m/s2 ) (5 kg + 5 kg) = 2.45 m/s2 . Since a > 0, m1 moves up the slope. Also, using equation (ii), T = m2 (g − a) = (5 kg)(7.36 m/s 2 ) = 36.8 N. T x m2 g N1 F21 N2 F m1g F12 m2 g Draw the free-body diagrams for each object. It is the contact force of m1 on m2 that causes m2 to accelerate. (a) Consider only horizontal forces (why??) ... #1 ... F − F21 = m1a. Question 4.11: Two blocks are in contact on a ! frictionless, 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) when F = 3.2 N, m1 = 2 kg and m2 = 6 kg. #2 ... F12 = m2a. But F12 = F21. ∴a = F . m1 + m2 The contact force is: F12 (= F21 ) = (b) a = m2 F. m1 + m2 F 3.2 = = 0.400 m/s2 . m1 + m2 2 + 6 F12 = 6 × 3.2 = 2.4 N. 2+6 ! F ! F (a) m1 < m2 (b) ! Do the results change if the force F is applied from the right? Conceptually, the magnitude of the acceleration ! will be the same because the total mass (m1 + m2 ) and F are the same. But, the contact force will be smaller since a smaller force will be required to produce the same acceleration for m1. We get the same result analytically. N1 F21 F12 F m1g We have N2 m2 g F − F12 = m2a, and F21 = m1a. Eliminating a, and putting F12 = F21, we find m1 F21 = F, (m1 + m2 ) i.e., the contact force of m2 on m1, F21, depends on the mass of the second block, m1, which in this case is smaller (2 kg) than in the problem (6 kg).