Dynamics: Newton s Laws of Motion

Transcription

1 Lecture 7 Chpter 4 Physics I Dynmics: Newton s Lws of Motion Solving Problems using Newton s lws Course website: Lecture Cpture:

2 Outline Chpter 4. Sections Types of forces ree-body digrms Solving Problem using Newton s lws

3 Applying Newton s Lws Know the forces : find the motion Know the motion : find the forces

4 Solving Problems with Newton s Lws: ree-body Digrms 1. Drw sketch. 2. or one object, drw free-body digrm, showing ll the forces cting on the object. Mke the mgnitudes nd directions s ccurte s you cn. Lbel ech force. If there re multiple objects, drw seprte digrm for ech one. 3. Resolve vectors into components. 4. Apply Newton s second lw to ech component. 5. Solve.

5 Possible forces cting on system Grvity g =mg Norml orce Tension Applied orce N Internl orces ( 12, 21 ) cncel ech other (N. 3 rd Lw)

6 ConcepTest 1 Going Up I A block of mss m rests on the floor of n elevtor tht is moving upwrd t constnt speed. Wht is the reltionship between the force due to grvity nd the norml force on the block? 1) N > mg 2) N = mg 3) N < mg (but not zero) 4) N = 0 5) depends on the size of the elevtor The block is moving t constnt speed, so it must hve no net force on it. The forces on it re N (up) nd mg (down), so N = mg, just like the block t rest on tble. m v

7 y Two blocks problem x Put it into (2) m 1 m 2 There re two objects, drw seprte digrm for ech one. m 1 m 2 Add them Internl forces cncel ech other!!!!!!!!!!!!! Cncel due to Newton s Third Lw 21 = m2 Given:, m 1, m 2 ind:, 12, 21 (contct forces between m 1 nd m 2 ) 21 = 12 x = m1 x 12 = m x = m m 2 x 21 = = m1 + m2 m = 2 = = 21 m1 + m2 m 1 + m 2 21 = 12 (1) (2)

8 Two blocks problem We cn now forget bout the internl forces y x m=m 1 +m 2 Given:, m 1, m 2 ind: (common ccelertion) Tret, m=m 1 +m 2, s the system (one big block) Apply N. 2 nd lw to m x component of N. 2 nd lw = m = = m + m 1 2 ( m + m ) 1 2

9 y Two blocks problem: y-eqution N m Box is described object x g =mg mg = y = 0 (no motion in y direction) N m y N = mg Y eqution gives norml force

10 Similr Problems Two boxes re held together by cble nd pulled by string on frictionless tble. y 2 Hnging buckets. mg t 2 t1 = 0 t t mg 1 y mg = 0 x m 1 m 2 mg

11 Block slides down n incline Choice of xes (friendly dvice) n Given: m, θ, no friction ind:, norml force Typiclly, equtions simplify if one xis chosen to be long observed ccelerted motion θ mg θ Replce forces with their components long chosen xes + mg cosθ = n m y y = 0 (no y-motion) n = mg cosθ

12 ConcepTest 2 Norml orce Below you see two cses: physics student pulling or pushing sled with force tht is pplied t n ngle θ. In which cse is the norml force greter? A) cse 1 B) cse 2 C) it s the sme for both D) depends on the mgnitude of the force E) depends on the ice surfce In cse 1, the force is pushing down (in Cse 1 ddition to mg), so the norml force needs to be lrger. In cse 2, the force is pulling up, ginst grvity, so the Cse 2 norml force is lessened.

13 Atwood mchine Two boxes of mss m 1 nd m 2 re held together by cble tht psses over mssless frictionless pulley. ind: ), ccelertion of ech box b), tension in cord Pulleys just redirect the tension The force of tension hs the sme mgnitude throughout the cble The mgnitude of the ccelertion is the sme for both objects They move in opposite directions m 1 m2

14 Atwood mchine or m 1 or m 2 m 1 g = m 1 m g = m ( 2 2 The mgnitude of the ccelertion is the sme for both objects (1)-(2) Using (1) 2 = 1 = m m ) g = ( m m ) ( ( m ( m = 2 1 m + m 1 2 ) ) g = m1 ( g + ) ) m 1 m 1 g Limit: if m 2 =0, then =g m 1 would be freely flling object If m 1 =m 2, then =0 (1) (2) m 2 m 2 g y

15 Another typicl problem A 5-kg block (m 2 ) sits on n inclined plne tilted t n ngle of 30º. It is ttched vi mssless cord to 2-kg block (m 2 ) lying on flt surfce. All surfces re frictionless. Drw free body digrm for ech block (include coordinte system). Wht re the mgnitude of the Norml orces on ech block? ind: ) ccelertion of the blocks; b) tension in the cord Eq. for m 1 y x 0 N 1 m1g = m1 y = m 1 y x m 1 N1 N2 m 2 g sss θ Eq. for m 2 y x 0 N 2 m2g cos θ = m2 y m2g sin θ t = m2 m 1 g θ m 2 g

16 Another typicl problem (cont.) N1 = m 1 g N 2 = m2g cosθ = m 1 m2g sin θ t = m2 Add them: = m 2 g sinθ = ( m1 + m2) m2g sinθ m + m 1 2 y (1) (2) x m 1 m 1 g N1 m m = m m + m = g 1 2 θ m 2 g N2 sinθ m 2 g sss θ

17 ConcepTest 3 Bowling vs. Ping-Pong I In outer spce, bowling bll nd Ping-Pong bll ttrct ech other due to grvittionl forces. How do the mgnitudes of these ttrctive forces compre? A) the bowling bll exerts greter force on the Ping-Pong bll B) the Ping-Pong bll exerts greter force on the bowling bll C) the forces re equl D) the forces re zero becuse they cncel out E) there re ctully no forces t ll The forces re equl nd opposite by Newton s Third Lw! 12 21

18 ConcepTest 4 Bowling vs. Ping-Pong II In outer spce, grvittionl forces exerted by bowling bll nd Ping-Pong bll on ech other re equl nd opposite. How do their ccelertions compre? The forces re equl nd opposite this is Newton s Third Lw!! But the A) they do not ccelerte becuse they re weightless B) ccelertions re equl, but not opposite C) ccelertions re opposite, but bigger for the bowling bll D) ccelertions re opposite, but bigger for the Ping-Pong bll E) ccelertions re equl nd opposite ccelertion is /m nd so the smller mss hs the bigger ccelertion.

19 Summry of Chpter 4 Newton s first lw: If the net force on n object is zero, it will remin either t rest or moving in stright line t constnt speed. Newton s second lw: Newton s third lw: Weight is the grvittionl force on n object. ree-body digrms re essentil for problem-solving. Do one object t time, mke sure you hve ll the forces, pick coordinte system nd find the force components, nd pply Newton s second lw long ech xis.

20 Thnk you See you on Mondy