CHAPTE 4: DYNAMICS: FOCE AND NEWTON S LAWS OF MOTION 4. NEWTON S SECOND LAW OF MOTION: CONCEPT OF A SYSTEM. A 6.- kg sprinter starts a race ith an acceleration of external force on him? 4. m/s. What is the net The net force acting on the sprinter is given b net F ma (6.kg)(4. m/s²) 65N 7. (a) If the rocket sled shon in Figure 4. starts ith onl one rocket burning, hat is its acceleration? Assume that the mass of the sstem is kg, and the force of friction opposing the motion is knon to be 65 N. (b) Wh is the acceleration not one- fourth of hat it is ith all rockets burning? 4 (a) Use the thrust given for the rocket sled in Figure 4.8, T.59 N. With onl one rocket burning, net F T f so that Neton s second la gives: a net F m T f m.59 4 N 65 N kg m/s (b) The acceleration is not one- fourth of hat it as ith all rockets burning because the frictional force is still as large as it as ith all rockets burning.. The eight of an astronaut plus his space suit on the Moon is onl 5 N. Ho much do the eigh on Earth? What is the mass on the Moon? On Earth? 6
Moon m g Earth mg mg Moon Moon Moon Earth 5 N.67 m/s 5 kg ( 5 kg)( 9.8 m/s ) 47 N.5 N Mass does not change. The astronaut s mass on both Earth and the Moon is 5 kg. 4.6 POBLEM- SOLVING STATEGIES 5. Calculate the force a 7.- kg high jumper must exert on the ground to produce an upard acceleration 4. times the acceleration due to gravit. Explicitl sho ho ou follo the steps in the Problem- Solving Strateg for Neton s las of motion. Step. Use Neton s Las of Motion. Step. Given: a 4. g (4.)(9.8 m/s ) 9. m/s ; m 7. kg Find F. Step. F + F ma, so that F ma + ma + mg m( a + g) F (7. kg)[(9. m/s ) + (9.8 m/s )].4 N The force exerted b the high- jumper is actuall don on the ground, but F is up from the ground to help him jump. Step 4. This result is reasonable, since it is quite possible for a person to exert a force 7
of the magnitude of N.. (a) Find the magnitudes of the forces F and F that add to give the total force F tot shon in Figure 4.5. This ma be done either graphicall or b using trigonometr. (b) Sho graphicall that the same total force is obtained independent of the order of addition of F and F. (c) Find the direction and magnitude of some other pair of vectors that add to give F tot. Dra these to scale on the same draing used in part (b) or a similar picture. (a) Since F is the - component of the total force: F F sin 5 ( N)sin5.47 N N. tot And F is the x - component of the total force: F F cos5 ( N)cos5 6.8 N 6 N. tot (b) F F F tot 5 is the same as: (c) For example, use vectors as shon in the figure. F tot 5 F ' F ' Fʹ is at an angle of from the horizontal, ith a magnitude of F cos ʹ F F 6.8 N Fʹ 7.4 N 7 N cos cos Fʹ is at an angle of 9 from the horizontal, ith a magnitude of F ʹ F Fʹ sin 5. N 8
. What force is exerted on the tooth in Figure 4.8 if the tension in the ire is 5. N? Note that the force applied to the tooth is smaller than the tension in the ire, but this is necessitated b practical considerations of ho force can be applied in the mouth. Explicitl sho ho ou follo steps in the Problem- Solving Strateg for Neton s las of motion. Step : Use Neton s las since e are looking for forces. Step : Dra a free bod diagram: Step : Given T 5. N, find F app. Using Neton s las gives Σ F, so that the applied force is due to the - components of the to tensions: ( 5. N) sin5.9 N F app T sinθ The x - components of the tension cancel. F x Step 4: This seems reasonable, since the applied tensions should be greater than the force applied to the tooth. 4. Figure 4.9 shos Superhero and Trust Sidekick hanging motionless from a rope. Superhero s mass is 9. kg, hile Trust Sidekick s is 55. kg, and the mass of the rope is negligible. (a) Dra a free- bod diagram of the situation shoing all forces acting on Superhero, Trust Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope beteen Superhero and Trust Sidekick. Indicate on our free- bod diagram the sstem of interest used to solve each part. 9
(a) (b) Using the upper circle of the diagram, F, so thatt ' T B. Using the loer circle of the diagram, F, giving T. Next, rite the eights in terms of masses: m g, m g. B B Solving for the tension in the upper rope gives: T ' T + + m g + m g g( m + m B B B Plugging in the numbers gives: T ' ( 9.8 m/s )( 55. kg + 9. kg).4 N Using the loer circle of the diagram, net F, so that T. Again, rite the eight in terms of mass: m g. Solving for the tension in the loer rope gives: T m g (55. kg) ( 9.8 m/s ) 59 N B ) 4
4.7 FUTHE APPLICATIONS OF NEWTON S LAWS OF MOTION 46. Integrated Concepts A basketball plaer jumps straight up for a ball. To do this, he loers his bod. m and then accelerates through this distance b forcefull straightening his legs. This plaer leaves the floor ith a vertical velocit sufficient to carr him.9 m above the floor. (a) Calculate his velocit hen he leaves the floor. (b) Calculate his acceleration hile he is straightening his legs. He goes from zero to the velocit found in part (a) in a distance of. m. (c) Calculate the force he exerts on the floor to do this, given that his mass is kg. (a) After he leaves the ground, the basketball plaer is like a projectile. Since he reaches a maximum height of.9 m, v v g( ), ith.9 m, and v m/s. Solving for the initial velocit gives: / / ( )] [(9.8 m/s )(.9 m)] 4. m/s v [g (b) Since e ant to calculate his acceleration, use v v + a( ), here. m, and since he starts from rest, v m/s. Solving for the v acceleration gives: a ( (4. m/s) 9.4 m/s ) ()(. m) (c) No, e must dra a free bod diagram in order to calculate the force exerted b the basketball plaer to jump. The net force is equal to the mass times the acceleration: net F ma F F mg So, solving for the force gives: F ma + mg m a g ( + ) kg(9.4 m/s + 9.8 m/s ) 4. N 4
49. Integrated Concepts An elevator filled ith passengers has a mass of 7 kg. (a) The elevator accelerates upard from rest at a rate of. m/s for.5 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upard at constant velocit for 8.5 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of.6 m/s for. s. What is the tension in the cable during deceleration? (d) Ho high has the elevator moved above its original starting point, and hat is its final velocit? (a) T m The net force is due to the tension and the eight: net F ma T T mg, and m 7 kg. a. m/s, sothe tension is : T m a 4 ( + g) (7 kg)(. m/s + 9.8 m/s ).87 N (b) a m/s, so the tension is: (7 kg)(9.8 m/s ) 67 4 T mg. N (c) a.6 m/s,but don : T m g a (7 kg)(9.8 m/s 4 ( ).6 m/s ).56 N (d) v t v t v t Use vt + at and v v + at. 4
For part (a), v m/s, a. m/s, t 5 s, given at (. m/s )(.5 s).5 m and v a t (. m/s )(.5 s).8 m/s. For part (b), v v.8 m/s, a m/s, t 8.5 s, so vt (.8 m/s)(8.5s) 5. m. For part (c), v.8 m/s, a.6 m/s, t. s, so that: v vt + a t v + a t (.8 m/s)(. s) +.5(.6 m/s)(. s).8 m/s + (-.6 m/s )(.s) m/s.7 m Finall, the total distance traveled is + +.5 m + 5. m +.7 m 9.5 m 9. 4 m And the final velocit ill be the velocit at the end of part (c), orv final m/s. 5. Unreasonable esults A 75.- kg man stands on a bathroom scale in an elevator that accelerates from rest to. m/s in. s. (a) Calculate the scale reading in netons and compare it ith his eight. (The scale exerts an upard force on him equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or hich premises are inconsistent? (a) Using v v + o at gives: v v. m/s m/s a 5. m/s. t. s No, using Neton s las gives net F F ma, so that 4
( + ) 75. kg( 5. m/s + 9.8 m/s ) 86 N F m a g. The ratio of the force to the eight is then: F m( a + g) mg 5. m/s + 9.8 m/s 9.8 m/s.5 (b) The value (86 N) is more force than ou expect to experience on an elevator. (c) The acceleration a 5. m/s. 5g is much higher than an standard elevator. The final speed is too large (. m/s is VEY fast)! The time of.s is not unreasonable for an elevator. 44