= 4.0 N. For the upper magnet, F L on U. T on U. because these are an action/ reaction pair. Equilibrium for the upper magnet requires n T on U

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1 7.8. (a) The upper magnet is labeled U, the lower magnet L. Each magnet eerts a long-range magnetic force on the other. Each magnet and the table eert a contact force (normal force) on each other. In addition, the table eperiences a normal force due to the surface. (b) Solve: Each object is in static equilibrium with ( F net ) 0. Start with the lower magnet. Because FU on L 3( FG ) L 6.0 N, equilibrium requires n 4.0 N. For the upper magnet, F L on U F U on L 6.0 N because these are an action/ reaction pair. Equilibrium for the upper magnet requires n T on U 8.0 N. For the table, action/reaction pairs are n L on T n 4.0 N and nu on T nt on U 8.0. The table s gravitational force is ( F ) 0 N, leaving n 4 N in order for the table to be in equilibrium. Summarizing, G T Upper magnet Table Lower magnet ( F G) U.0 N ( F G) T 0 N ( F G) L.0 N n T on U 8.0 N n U on T 8.0 N n 4.0 N F L on U 6.0 N n L on T 4.0 N F U on L 6.0 N n 4 N Assess: The result n 4 N makes sense. The combined gravitational force on the table and two magnets is 4 N. Because the table is in equilibrium, the upward normal force of the surface has to eactly balance the total gravitational force on the table and magnets.

2 7.10. Model: The blocks are to be modeled as particles and denoted as 1,, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. The force applied on block 1 is F A on 1 1 N. The acceleration for all the blocks is the same and is denoted by a. Solve: (a) Newton s second law for the three blocks along the -direction is ( Fon 1) FA on 1 F on 1 ma 1 ( Fon ) F1on F3 on ma ( ) F F m a on 3 on 3 3 Adding these three equations and using Newton s third law ( F on 1 F1 on and F3 on F on 3), we get ( + + ) ( ) ( + + ) F m m m a Aon1 1 3 Using this value of a, the force equation on block 3 gives (b) Substituting into the force equation on block 1, 1 N 1 kg kg 3 kg a a m/s Fon3 m3a 3 kg m/s 6 N on1 1 N F 1 kg m/s Fon1 10 N Assess: Because all three blocks are pushed forward by a force of 1 N, the value of 10 N for the force that the kg block eerts on the 1 kg block is reasonable.

3 7.1. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley. The system is the man and the block. Solve: Clearly the entire system remains in equilibrium since m. B > m M The block would move downward but it is already on the ground. From the free-body diagrams, we can write down Newton s second law in the vertical direction as ( Fon M ) TR on M ( FG ) 0 N T ( F ) ( M )( ) y Since the tension is the same on both sides, TBonR TMonR T 588 N. RonM G 60 kg 9.8 m/s 588 N M

4 7.14. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as objects in the system. Solve: The force F et acts only on the rope. Since the rope and the ice block move together, they have the same acceleration. Also because the rope has mass, F et on the front end of the rope is not the same as FI on R that acts on the rear end of the rope. Newton s second law along the -ais for the ice block and the rope is ( F ) F ma on I R on I I 10 kg.0 m/s 0 N ( ) Fon R F et FI on R mra et R on I R F F m a Fet FR on I + mra 0 N kg.0 m/s 1 N

5 7.18. Please refer to Figure P7.18. Solve: Since the ropes are massless we can treat the tension force they transmit as a Newton s third law force pair on the blocks. The connection shown in figure P7.18 has the same effect as a frictionless pulley on these massless ropes. The blocks are in equilibrium as the mass of A is increased until block B slides, which occurs when the static friction on B is at its maimum value. Applying Newton s first law to the vertical forces on block n F m g The static friction force on B is thus B gives ( G). B B B ( fs) μ B snb μsmbg. Applying Newton s first law to the horizontal forces on B gives ( fs) T B A on B, vertical forces on A gives T ( F ) m g Since T A on B T, B on A we have ( f ) B on A G A A. μ mg mg s B A ma μsmb kg 1.0 kg and the same analysis of the s mg B A, so