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 (E). In indicating the various forces, we have denoted the normal (contact) forces b n r, the kinetic frictional (contact) forces b f r, and the weights b w r. All the action/reaction forces have been identified with dotted lines. The bottom figure shows the free bod diagrams of each block, the rope, and the pulle.
8.9. Model: The blocks are to be modeled as particles and denoted as,, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three sstems of interest. The force applied on block is F A on = N. The acceleration for all the blocks is the same and is denoted b a. Solve: Newton s second law for the three blocks along the -direction is ( Fon ) = FA on F on = ma ( Fon ) = Fon F3 on = ma F = F = m a on 3 on 3 3 Adding these three equations and using Newton s third law (F on = F on and F 3 on = F on 3 ), we get = ( + + ) = ( + + ) F m m m a A on 3 Using this value of a, the force equation on block 3 gives Substituting into the force equation on block, N kg kg 3 kg a a = m/s F on 3 = m3a = 3 kg m/s = 6 N on N F = kg m/s F on = 0 N Assess: Because all three blocks are pushed forward b a force of N, the value of 0 N for the force that the kg block eerts on the kg block is reasonable.
8.4. Model: The hanging block and the rail car are separate sstems. Solve: The mass of the rope is ver small in comparison to the 000 kg block, so we will assume a massless rope. In that case, the forces T r r and T act as if the are an action/reaction pair. The hanging block is in static r equilibrium, with F net = 0 N, so T = mblockg = 9,600 N. The rail car with the pulle is also in static equilibrium: T + T3 T = 0 N Notice how the tension force in the cable pulls both the top and bottom of the pulle to the right. Now, T = T = 9,600 N b Newton s third law. Also, the cable tension is T = T3 = T. Thus, T = T = 9800 N. 3
8.0. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction. a A = a B = a. Newton s second law on sled A is F r on A = na wa = 0 N na wa F r = T f = m a Solve: The acceleration constraint is Using f A = µ k n A, the -equation ields On sled B: T µ n m a on A k A A = = ( ) = F r on B = nb wb = 0 N nb wb on A on A A A 50 N 0. 00 kg 9.8 m/s 00 kg a a = 0.5 m/s = T on B and T on A act as if the are an action/reaction pair, so on B 50 N. ( 9.8 m/s ) = 78.4 N, we get Thus the tension T = 70 N. F r = T T f = m a on B on B B B T 50 N 78.4 N = 80 kg 0.5 m/s T = 70 N T = Using f = µ n = B k B 0.0 80 kg 4
8.. The car and the ground are denoted b C and G, respectivel. Solve: (a) As the drive wheels turn the push backward against the ground. This is a static friction force F r C on G because the wheels don t slip against the ground. B Newton s third law, the ground eerts a reaction force F r G on C. This reaction force is opposite in direction to F r C on G, hence, is in the forward direction. This is the force that accelerates the car. (b) The car has an internal source of energ fuel that allows it to turn the wheels and eert the force This is an active push b the car, generating the force F r G on C F r C on G. in response. Houses do not have an internal source of energ that allows them to push sidewas against the ground. (c) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although that is not critical) and the nondrive wheels. For this car, two-thirds of the weight rests on the front wheels. Phsicall, force F r G on C is a static friction force. The maimum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches its maimum possible value. = = µ = µ ( 3 ) F f n w G on C s ma s F s F =.00 500 kg 9.8 m/s = 9800 N F 9800 N ama = = = 6.53 m/s m 500 kg G on C 5
8.5. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constant-acceleration kinematic equations. Solve: Package B has a smaller coefficient of friction. It will tr to overtake package A and push against it. a = a = a Package A will push back on B. The acceleration constraint is Newton s second law for each package is A B. F = F + w sinθ f = m a on A B on A A ka A ( cos ) F + m gsinθ µ m g θ = m a B on A A ka A A F = F f + w sinθ = m a on B A on B kb B B ( cos ) F µ m g θ + m gsinθ = m a A on B kb B B B where we have used na = ma cos θ g and nb = mb cos θ g. Adding the two force equations, and using FA on B = FB on A because the are an action/reaction pair, we get a = gsinθ Finall, using v ( t t ) a( t t ) = + +, 0 0 0 0 ( µ m + µ m )( gcosθ ) ka A kb B m A + m B ( t ) =.87 m/s m = 0 m + 0 m +.87 m/s 0 s t =.48 s 6
8.8. Model: Blocks and are our sstems of interest and will be treated as particles. Assume a frictionless rope and massless 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 free-bod diagrams, especiall force f r eerted on block b block. We have quite a few pieces of information to include. First, Newton s second law for blocks and : r F = f T = µ n T = m a = m a F = n m g = 0 N n = m g net on k net on F = T f f T = T f µ 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 = mg f = f = µ kn = µ kmg T = T = 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: µ = µ µ kmg T ma Subtract the first equation from the second to get T T m g m + m g = m a pull k k T µ 3m + m g Tpull µ k ( 3m + m ) g = ( m + m ) a a = =.77 m/s m + m pull k 7
8.30. Model: Masses m and m are considered particles. The string is assumed to be massless. Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it acts like a pulle. Thus tension forces T r and T r act as if the were an action/reaction pair. Mass m is in circular motion of radius r, so Newton s second law for m is m v Fr = T = r Mass m is at rest, so the -equation of Newton s second law is F = T m g = 0 N T = m g Newton s third law tells us that T = T. Equating the two epressions for these quantities: m v m rg = mg v = r m 8
8.34. Model: Use the particle model for the block of mass M and the two massless pulles. Additionall, the rope is massless and the pulles are frictionless. The block is kept in place b an applied force F r. Solve: Since there is no friction on the pulles, T = T 3 and T = T 5. Newton s second law for mass M is T w = 0 N T = Mg = 0. kg 9.8 m/s = 00 N Newton s second law for the small pulle is T T + T3 T = 0 N T = T3 = = 50 N = T5 = F Newton s second law for the large pulle is T T T T = 0 N T = T + T + T = 50 N 4 3 5 4 3 5 9
8.35. Model: Assume the particle model for m, m, and m 3, 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 w = m a. Newton s second law for m is Newton s second law for m 3 is ( Fon m ) n w n ( m ) = = 0 N = kg 9.8 m/s = 9.6 N F = T f T = m a T µ n T = ( kg) a on k ( m ) 3 F = T w = m a on 3 3 3 k Since m, m, and m 3 move together, a = a = a 3 = a. The equations for the three masses thus become = = T µ n T = m a = a = = T w ma kg a k kg Subtracting the third equation from the sum of the first two equations ields: w µ kn + w3 = 6 kg a T w3 m3a 3 kg a kg 9.8 m/s 0.300 9.6 N + 3 kg 9.8 m/s = 6 kg a a =.9 m/s 0
8.40. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulles. Note that for ever meter block moves forward, one meter is provided to block. So each rope on m has to be lengthened b one-half meter. Thus the acceleration constraint is a = a. Solve: Newton s second law for block is T = m a. Newton s second law for block is T w = m a. Combining these two equations gives where we have used a = a. Assess: mg ( ma ) mg = m ( a ) a 4m + m = mg a = 4m + m If m = 0 kg, then a = g. This is what is epected for a freel falling object. 8.44. Model: Use the particle model for the wedge and the block. The block will not slip relative to the wedge if the both have the same acceleration a. Solve: The -component of Newton s second law for block m is m g cosθ ( Fon ) = n cos θ w = 0 N n = Combining this equation with the -component of Newton s second law ields: Now, Newton s second law for the wedge is n sinθ F = n = m a a = = gtanθ m ( on ) sinθ F = F n sinθ = m a on F = m a + n sinθ = m a + m a = (m + m ) a = (m + m ) g tanθ