Chapter 5 Force and Motion I

Transcription

1 Chapter 5 orce and Motion I I. ewton s first law. II. ewton s second law. III. Particular forces: -Gravitational - Weiht - ormal - riction - ension IV. ewton s third law. ewton mechanics laws cannot be applied when: ) he speed of the interactin bodies are a fraction of the speed of liht Einstein s special theor of relativit. ) he interactin bodies are on the scale of the atomic structure Quantum mechanics I. ewton s first law: If no net force acts on a bod, then the bod s velocit cannot chane; the bod cannot accelerate v = constant in manitude and direction. - Principle of superposition: when two or more forces act on a bod, the net force can be obtained b addin the individual forces vectoriall. - Inertial reference frame: where ewton s laws hold.

2 II. ewton s second law: he net force on a bod is equal to the product of the bod s mass and its acceleration. net = ma (5.) net x = max, net, = ma, net, z = ma, z (5.) - he acceleration component alon a iven axis is caused onl b the sum of the force components alon the same axis, and not b force components alon an other axis. - Sstem: collection of bodies. - External force: an force on the bodies inside the sstem. III. Particular forces: -Gravitational: pull directed towards a second bod, normall the Earth = m (5.3) - Weiht: manitude of the upward force needed to balance the ravitational force on the bod due to an astronomical bod W = m (5.4) - ormal force: perpendicular force on a bod from a surface aainst which the bod presses. = m (5.5) - rictional force: force on a bod when the bod attempts to slide alon a surface. It is parallel to the surface and opposite to the motion. -ension: pull on a bod directed awa from the bod alon a massless cord.

3 IV. ewton s third law: BC = CB When two bodies interact, the forces on the bodies from each other are alwas equal in manitude and opposite in direction. (5.6) QUESIOS Q. wo horizontal forces, pull a banana split across a frictionless counter. Without usin a calculator, determine which of the vectors in the free bod diaram below best represent: a), b). What is the net force component alon (c) the x-axis, (d) the -axis? Into which quadrant do (e) the net-force vector and (f) the split s acceleration vector point? i ˆ = (3 )ˆ (4 ) j = ( )ˆ i () ˆj net = + = ()ˆ i (6) ˆj Same quadrant, 4 I. rictional force Counter force that appears when an external force tends to slide a bod alon a surface. It is directed parallel to the surface and opposite to the slidin motion. -Static: (f s ) compensates the applied force, the bod does not move. = f s // o motion -Kinetic: (f k ) appears after a lare enouh external force is applied and the bod loses its intimate contact with the surface, slidin alon it. (applied force) Acceleration Constant velocit 3

4 f < f = (6.) k f s,max s, max µ s If fs bod slides // >, max riction coefficients f = k µ k (6.) After the bod starts slidin, f k decreases. Q. he fiure below shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If the force manitudes are chosen properl, in which situation it is possible that the block is (a) stationar and (b) movin with constant velocit? a 0 a=0 a=0 a 0 net Q5. In which situations does the object acceleration have (a) an x-component, (b) a component? (c) ive the direction of a. net 4

5 Q. A bod suspended b a rope has a weih of 75. Is equal to, reater than, or less than 75 when the bod is movin downward at (a) increasin speed and (b) decreasin speed? net = = ma = m( a) Movement (a) Increasin speed: v f >v 0 a>0 < (b) Decreasin speed: v f < v 0 a<0 > Q8. he fiure below shows a train of four blocks bein pulled across a frictionless floor b force. What total mass is accelerated to the riht b (a), (b) cord 3 (c) cord? (d) Rank the blocks accordin to their accelerations, reatest first. (e) Rank the cords accordin to their tension, reatest first. 3 (a) pulls m total = (0+3+5+)k = 0k (b) Cord 3 3 m=(0+3+5)k = 8k (c) Cord m= 0k (d) =ma All tie, same acceleration (e) - 3 = a 3 - = 5a - =3a =0a - 3 = a =8a+a=0a 3-3a= 5a 3 =8a -0a=3a =3a =0a Q. A to box is on top of a heavier do house, which sits on a wood floor. hese objects are represented b dots at the correspondin heihts, and six vertical vectors (not to scale) are shown. Which of the vectors best represents (a) the ravitational force on the do house, (b) on the to box, (c) the force on the to box from the do house, (d) the force on the do house from the to box, (e) force on the do house from the floor, (f) the force on the floor from the do house? () Which of the forces are equal in manitude? Which are (h) reatest and (i) least in manitude? (a) on do house: 4 or 5 (h) Greatest: 6,3 (b) on to box: (c) to from do house: (d) do-house from to box: 4 or 5 (e) do-house from floor: 3 (f) floor from do house: 6 () Equal: =, =5, 3=6 (i) Smallest:,,5 5

6 5. here are two forces on the k box in the overhead view of the fiure below but onl one is shown. he fiure also shows the acceleration of the box. ind the second force (a) in unit-vector notation and as (b) manitude and (c) direction. ˆ a = (cos 40 i + sin 40 ˆ) j m = ( 6ˆ i 0.39 ˆ) j m = ma = k( 6ˆ i 0.39 ˆ) j m = ( ˆ i 0.78 ˆ) j = + = 0ˆ i + x = = = 0.78 = = x = 38.7 x = 3 = ( 3ˆ i 0.78 ˆ) j 0.78 tanθ = = 33 3 or = 3 Rules to solve Dnamic problems - Select a reference sstem. - Make a drawin of the particle sstem. - Isolate the particles within the sstem. - Draw the forces that act on each of the isolated bodies. - ind the components of the forces present. - Appl ewton s second law (=ma) to each isolated particle. 6

7 9. (a) A k salami is supported b a cord that runs to a sprin scale, which is supported b another cord from the ceilin. What is the readin on the scale, which is marked in weih units? (b) Here the salami is supported b a cord that runs around a pulle and to a scale. he opposite end of the scale is attached b a cord to a wall. What is the readin on the scale? (c) he wall has been replaced b a second salami on the left, and the assembl is stationar. What is the readin on the scale now? W = = m = (k )(9.8m ) = 07.8 ( a) a = 0 = = 07.8 ( b) a = 0 = = ( c) a = 0 = = In all three cases the scale is not acceleratin, which means that the two cords exert forces of equal manitude on it. he scale reads the manitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in manitude as the weih of the salami because the salami is not acceleratin. 7

8 3. An electron with a speed of.x0 7 m/s moves horizontall into a reion where a constant vertical force of 4.5x0-6 acts on it. he mass of the electron is m=9.x0-3 k. Determine the vertical distance the electron is deflected durin the time it has moved 30 mm horizontall. d v 0 d x =0.03m net net d x = vxt = 0.03m = (. 0 m ) t t =. 4ns = ma = = ( k) a = a 7 6 (9. 0 = d = vot + 0.5at = 0.5 ( m ) (.5 0 s) = m m k)(9.8m 9 ) 3. In the fiure below, m block =8.5k and θ=30º. ind (a) ension in the cord. (b) ormal force actin on the block. (c) If the cord is cut, find the manitude of the block s acceleration. x ( a) a = 0 a = x x = 0 = m sin 30 = (8.5k)(9.8m )0.5 = 4.65 ( c) = 0 x = ma = 4.65 = 8.5a a = 4.9m x = 0 ( b) a = 0 = 0 = = m cos30 =

9 55. he fiure below ives as a function of time t, the force component x that acts on a 3k ice block, which can move onl alon the x axis. At t=0, the block is movin in the positive direction of the axis, with a speed of 3m/s. What are (a) its speed and (b) direction of travel at t=s? t = 0 v 0 t = s v = 3m f =? s x dvx dvx x ax = = dt = v f v0 = dt m dt dt m s otal raph area = 5s = xdt = ( v f v0) m = ( v f 3m )3k 5km v f = + 3m = 8m 3k 0 0 Midterm_extra_Sprin04. wo bodies, m= k and m=k are connected over a massless pulle. he coefficient of kinetic friction between m and the incline is 0.. he anle θ of the incline is 0º. Calculate: (a) Acceleration of the blocks. (b) ension of the cord., x = = m sin 0 = 6.7 = m cos 0 = 8.4 f = µ = µ m cos 0 =.84 k, Block : m = m a k 9.8 = a 0º f m m m m Block : f, x = m a = a Addin 3a =.6 a = 0.4m, = 9.38 Midterm_Sprin04. he three blocks in the fiure below are connected b massless cords and pulles. Data: m =5k, m =3k, m 3 =k. Assume that the incline plane is frictionless. (i) Show all the forces that act on each block. (ii) Calculate the acceleration of m, m, m 3. (iii) Calculate the tensions on the cords. (iv) Calculate the normal force actin on m =m cos30º x =m sin30º m m 3 30º m 3 m x m m Block : m - =m a Block : m (sin30º) + - =m a Block 3: -m 3 = m 3 a (i) Addin ()+()+(3) (m +0.5m -m 3 )=a(m +m +m 3 ) a= 4.4m/s (ii) =m (-a)= 5k(9.8 m/s -4.4 m/s ) = 6.95 (iii) =m 3 (+a)= k(9.8 m/s +4.4 m/s )= 8.4 (iv) = = m cos30º =

10 B. (a) What should be the manitude of in the fiure below if the bod of mass m=0k is to slide up alon a frictionless incline plane with constant acceleration a=.98 m/s? (b) What is the manitude of the ormal force? m( a + 0.5) cos 0 m sin 30 = ma = = 73. cos 0 m cos 30 sin 0 = 0 = º 0º B. Given the sstem plotted below, where m =k and m =6k, calculate the force necessar to lift up m with a constant acceleration of 0.m/s. he pulles and cords are massless, and the table surface is frictionless. x Movement m m d = at d d = = a t m = m a 0.5a t = a t a = 0.5a = 0.m a = 0.4m = = 0.5m ( a + ) = 0.5(6k)( ) m 30 m m = m a = + m a = 30 + (k)(0.4m ) = 0