Then, by Newton s third law: The knots are also in equilibrium. Newton s law applied to the left knot is. The y-equation gives T1 m1 g sin 1.

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1 Chapter 7 Solutions

2 7.7. Model: The two hanging blocks, which can be modeled as particles, together with the two knots where rope meets with rope and rope meets with rope form a sstem. All the four objects in the sstem are in static equilibrium. The ropes are assumed to be massless. Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are in static equilibrium with Fnet 0 N. Note that there are three action/reaction pairs. For Block and Block, Fnet 0 N and we have T F m g T F m g net 4 G 5 G Then, b Newton s third law: T4 T 4 m g T5 T 5 mg The knots are also in equilibrium. Newton s law applied to the left knot is F T T cos 0 N F T sin T T sin m g 0 N net 4 The -equation gives T m g sin. Substitute this into the -equation to find Newton s law applied to the right knot is net m g cos m g T sin tan F T cos T 0 N F T sin T T sin m g 0 N net 5 These can be combined just like the equations for the left knot to give mg cos mg T sin tan But the forces T and T are an action/reaction pair, so T T. Therefore,

3 7.4. Model: The coffee mug (M) is the onl object in the sstem, and it will be treated as a particle. The model of friction and the constant-acceleration kinematic equations will also be used. Solve: The mug and the car have the same velocit. If the mug does not slip, their accelerations will also be the v v a, we get same. Using 0 0 The static force needed to stop the mug is The maimum force of static friction is Since fs ma fs ma 0 m /s 0 m/s a 50 m F f ma 4.0 m/s net s 0.5 kg 4.0 m/s.0 N fs.0 N ( f ) n F mg kg 9.8 m/s.45 N s ma s s G s ( ) ( ), the mug does not slide. a

4 7.8. Model: The bo (B) and the crate (C) are the two objects in our sstem, and the will be treated in the particle model. We will also use the static and kinetic friction models. Solve: The fact that the bo s feet occasionall slip means that the maimum force of static friction must eist between the bo s feet and the sidewalk. That is, f n. Also f n. sb sb B kc kc C Newton s second law for the crate is Fon C nc FG 0 N n C C mcg Newton s second law for the bo is Fon B nb FG 0 N n B B mbg F C on B and F B on C are an action/reaction pair, so Fon C FB on C fkc 0 N FB on C fkc kcnc kcmc g Fon B fsb FCon B 0 N FC on B fsb sbnb sbmbg sbm kg B FC on B FB on C sbmbg kcmcg mc.0 0 kg 0. kc

5 7.0. Model: The two blocks form a sstem of interacting objects. Please refer to Figure P7.0. Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newton s first law applied to the -direction on Block L ields Therefore F 0 n F cos0 n m g cos0 L L G L L L fk k mlg cos kg 9.80 m/s cos0.84 N L L A similar analsis of the vertical forces on Block R gives fk.84 N as well. Using Newton s second law in R the -direction for Block L, F L mla T R on L fk F G sin0 m L L La TR on L.84 N mlg sin0. For Block R, FR mra F sin0.84 N G T R L on R mra mr gsin0.84 N TL on R These are two equations in the two unknowns a and TL on R TR on L T. Solving them, we obtain a. m/s T 0.6 N. Assess: The tension in the string is positive, and is about / of the kinetic friction force on each of the blocks, which is reasonable.

6 7.. Model: The two blocks ( and ) are the sstems of interest and will be treated as particles. The ropes are assumed to be massless, and the model of kinetic friction will be used. Solve: (a) The separate free-bod diagrams for the two blocks show that there are two action/reaction pairs. Notice how block both pushes down on block (force n ) and eerts a retarding friction force f top on the top surface of block. Block is in static equilibrium ( a 0 m/s ) but block is accelerating. Newton s second law for block is F f T 0 N T f net on rope rope F n m g 0 N n m g net on Although block is stationar, there is a kinetic force of friction because there is motion between block and block. The friction model means f. kn km g Substitute this result into the -equation to get the tension in the rope: (b) Newton s second law for block is F T f f a a m m net on pull top bot Trope f km g.9 N Fnet on n n mg a 0 m/s m m Forces n and n are an action/reaction pair, so n. n m g Substituting into the -equation gives n m m g This is not surprising because the combined weight of both objects presses down on the. surface. The kinetic friction on the bottom surface of block is then f n m m g bot k k The forces f and f top are an action/reaction pair, so f bot f. m g k Inserting these friction results into the -equation gives

7 7.. Model: The -kg and 4-kg blocks are to be treated as particles. The models of kinetic and static friction and the constant-acceleration kinematic equations will be used. Solve: Minimum time will be achieved when static friction is at its maimum possible value. Newton s second law for the 4-kg block is F n on 4 on 4 F G 0 N n 4 on 4 FG m4g f ( f ) n N.5 N Newton s second law for the -kg block is s4 s ma s on kg 9.8 m/s 9. N 4 Fon n n4on FG 0 N n n F Friction forces f and f s4 are an action/reaction pair. Thus Fon f s fk ma f s4 kn ma 4on G 9. N.0 kg 9.8 m/s 68.6 N.5 N N.0 kga Since block does not slip, this is also the acceleration of block 4. The time is calculated as follows: t v t t a t t a.67 m/s 5.0 m 0 m 0 m.67 m/s 0 s t.75 s

8 7.4. Model: Blocks and make up the sstem of interest and will be treated as particles. Assume a massless rope and frictionless pulle. Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a, a a where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if the are action/reaction pairs because we are assuming a massless rope and a frictionless pulle. Make sure ou understand wh the friction forces point in the directions shown in the freebod diagrams, especiall force f eerted on block b block. We have quite a few pieces of information to include. First, Newton s second law for blocks and : F f T n T m a m a net on k F n m g 0 N n m g net on F T f f T T f n T m a m a net on pull pull k F n n m g 0 N n n m g net on We ve alread used the kinetic friction model in both -equations. Net, Newton s third law: n n m g f f kn km g TT T Knowing n, we can now use the -equation of block to find n. Substitute all these pieces into the two - equations, and we end up with two equations in two unknowns: km g T ma Subtract the first equation from the second to get T T m g m m g m a pull k k T m m g Tpull k m m g m m a a.77 m/s m m pull k

9 7.8. Model: Assume the particle model for m, m, and m, and the model of kinetic friction. Assume the ropes to be massless, and the pulles to be frictionless and massless. Solve: Newton s second law for m is T ( FG ) ma. Newton s second law for m is Fon m n FG n Newton s second law for m is 0 N.0 kg 9.8 m/s 9.6 N ( F ) T f T m a T n T (.0 kg) a on m k k m F T F m a on G Since m, m, and m move together, a a a a. The equations for the three masses thus become T n T m a a T FG m a.0 kg a k.0 kg Subtracting the third equation from the sum of the first two equations ields: FG kn FG 6.0 kg a T FG m a.0 kg a.0 kg 9.8 m/s N.0 kg 9.8 m/s 6.0 kg a a. m/s

PHYS1100 Practice problem set, Chapter 8: 5, 9, 14, 20, 22, 25, 28, 30, 34, 35, 40, 44

PHYS1100 Practice problem set, Chapter 8: 5, 9, 14, 20, 22, 25, 28, 30, 34, 35, 40, 44 PHYS00 Practice problem set, Chapter 8: 5, 9, 4, 0,, 5, 8, 30, 34, 35, 40, 44 8.5. Solve: The top figure shows the pulle (P), block A, block B, the surface S of the incline, the rope (R), and the earth