17. (a) The two crates must have the same acceleration. From the force diagram for crate 1 we have x-component: F - F_{12}- u_{k}F_{N1}= m_{1}a; y-component: F_{N1}- m_{1}g = 0, or F_{N1}= m_{1}g. From the force diagram for crate 2 we have x-component: F_{12}- u_{k}F_{N2}= m_{2}a; y-component: F_{N2}- m_{2}g = 0, or F_{N2}= m_{2}g. If we add the two x-equations, we get F - u_{k}m_{1}g - u_{k}m_{2}g = m_{1}a + m_{2}a; F - u_{k}(m_{1}+ m_{2})g = (m_{1}+ m_{2})a; 750 N - (0.12)(80 kg + 210 kg)(9.80 m/s^{2}) = (80 kg + 210 kg)a, which gives a = 1.4 m/s^{2}. (b) We can find the force between the crates from the x-equation for crate 2: F_{12}- u_{k}m_{2}g = m_{2}a; F_{12}- (0.12)(210 kg)(9.80 m/s^{2}) = (210 kg)(1.41 m/s^{2}), which gives F_{12}= 5.4 x 10^{2}N. (c) If the crates are reversed, the acceleration will be the same: a = 1.4 m/s^{2}. For the force between the crates, we have F_{12}- u_{k}m_{1}g = m_{1}a; F_{12}- (0.12)(80 kg)(9.80 m/s^{2}) = (80 kg)(1.41 m/s^{2}), which gives F_{12}= 2.1 x 10^{2}N.

21. We choose the coordinate system shown in the force diagram and assume the cord is taut. (a) From the force diagram for each block, we have Sum(F) = ma: y-component (1) F_{N1}- m_{1}g cosq = 0, or F_{N1}= m_{1}g cosq; x-component (1): m_{1}g sinq - m_{1}F_{N1}- F_{T}= m_{1}a; (m_{1}sinq - m_{1}m_{1}cosq)g - F_{T}= m_{1}a; y-component (2) F_{N2}- m_{2}g cosq = 0, or F_{N2}= m_{2}g cosq; x-component (2): m_{2}g sinq - m_{2}F_{N2}+ F_{T}= m_{2}a; (m_{2}sinq - m_{2}m_{2}cosq)g + F_{T}= m_{2}a. When we add the two x-equations, we have (m_{1}+ m_{2})a = [(m_{1}+ m_{2}) sinq - (m_{1}m_{1}+ m_{2}m_{2}) cosq]g; (5.0 kg + 5.0 kg)a = {[(5.0 kg + 5.0 kg) sin 30^{o}- [(0.20)(5.0 kg) + (0.30)(5.0 kg)] cos 30^{o}}(9.80 m/s^{2}) which gives a = 2.8 m/s^{2}. (b) We find the tension from F_{T}= m_{1}[(sinq - m_{1}cosq)g - a] = (5.0 kg){[sin 30^{o}- (0.20) cos 30^{o}](9.80 m/s^{2}) - (2.78 m/s^{2})} = 2.1 N. Note that the positive result justifies our assumption that the cord is taut.

27. The direction of the kinetic friction force is determined by the direction of the velocity, not the direction of the acceleration. We assume the block on the plane is moving up, so the friction force is down. The forces and coordinate systems are shown in the diagram. From the force diagram, with the block m_{2}as the system, we can write Sum(F) = ma: y-component: m_{2}g - F_{T}= m_{2}a. From the force diagram, with the block m_{1}as the system, we can write Sum(F) = ma: x-component: F_{T}- F_{fr}- m_{1}g sinq = m_{1}a; y-component: F_{N}- m_{1}g cosq = 0; with F_{fr}= u_{k}F_{N}. When we eliminate F_{T}between these two equations, we get a = (m_{2}- m_{1}sinq - u_{k}m_{1}cosq)g/(m_{1}+ m_{2}). (a) For a mass m_{1}= 5.0 kg, we have a = [5.0 kg - (5.0 kg) sin 30^{o}- (0.10)(5.0 kg) cos 30^{o}](9.80 m/s^{2})/(5.0 kg + 5.0 kg) = 2.0 m/s^{2}up the plane. The acceleration is up the plane because the answer is positive. This agrees with our assumption for the direction of motion. (b) For a mass m_{1}= 2.0 kg, we have a = [5.0 kg - (2.0 kg) sin 30^{o}- (0.10)(2.0 kg) cos 30^{o}](9.80 m/s^{2})/(2.0 kg + 5.0 kg) = 5.4 m/s^{2}up the plane. The acceleration is up the plane because the answer is positive. This agrees with our assumption for the direction of motion.

41. The static friction force provides the centripetal acceleration. We write Sum(F) = ma from the force diagram for the coin: x-component: F_{fr}= mv^{2}/R; y-component: F_{N}- mg = 0. The highest speed without sliding requires F_{fr},max = u_{s}F_{N}. The maximum speed before sliding is v_{max}= 2 pi R/Tmin = 2 pi Rfmax = 2 pi (0.120 m)(50 /min)/(60 s/min) = 0.628 m/s. Thus we have u_{s}mg = mv_{max}^{2}/R u_{s}(9.80 m/s^{2}) = (0.628 m/s)^{2}/(0.120 m), which gives u_{s}= 0.34.

84. We assume there is no tension in the rope and simplify the forces to those shown. From the force diagram, we have Sum(F) = ma: x-component: F_{N}shoes - F_{N}wall = 0, so the two normal forces are equal: F_{N}shoes = F_{N}wall = F_{N}; y-component: F_{fr}shoes + F_{fr}wall - mg = 0. For a static friction force, we know that Fsfr ~ u_{s}F_{N}. The minimum normal force will be exerted when the static friction forces are at the limit: u_{s}shoesF_{N}shoes + u_{s}wallF_{N}wall = mg; (0.80 + 0.60)F_{N}= (70 kg)(9.80 m/s^{2}), which gives F_{N}= 4.9 x 10^{2}N.

86. (a) If motion is just about to begin, the static friction force on the block will be maximum: F_{fr},max = u_{s}F_{N}, and the acceleration will be zero. We write Sum(F) = ma from the force diagram for each object: x-component (block): F_{T}- u_{s}F_{N}= 0; y-component (block): F_{N}- m_{1}g = 0; y-component (bucket): (m_{2}+ m_{sand})g - F_{T}= 0. When we combine these equations, we get (m_{2}+ m_{sand})g = u_{s}m_{1}g, which gives m_{sand}= u_{s}m_{1}- m_{2}= (0.450)(28.0 kg) - 1.00 kg = 11.6 kg. (b) When the system starts moving, the friction becomes kinetic, and the tension changes. The force equations become x-component (block): F_{T2}- u_{k}F_{N}= m_{1}a; y-component (block): F_{N}- m_{1}g = 0; y-component (bucket): (m_{2}+ m_{sand})g - F_{T2}= (m_{2}+ m_{sand})a. When we combine these equations, we get (m_{2}+ m_{sand}- u_{k}m_{1})g = (m_{1}+ m_{2}+ m_{sand})a, which gives a = [1.00 kg + 11.6 kg - (0.320)(28.0 kg)](9.80 m/s^{2})/(28.0 kg + 1.00 kg + 11.6 kg) = 0.879 m/s^{2}.

18. (a) While the block is sliding down, friction will be up the plane, opposing the motion. From the force diagram for the block, we have Sum(F) = ma: x-component: mg sinq - u_{k}F_{N}= ma. y-component: F_{N}- mg cosq = 0. When we combine these, we have a = g(sinq - u_{k}cosq) = (9.80 m/s^{2})[sin 22.0^{o}- (0.17) cos 22.0^{o}] = 2.1 m/s^{2}. (b) For the motion of the block, we have v^{2}= v_{0}^{2}+ 2a(x - x0) = 0 + 2(2.13 m/s^{2})(9.3 m), which gives v = 6.3 m/s.

24. (a) For the hanging box we can write Sum(F) = ma: y-component: m_{2}g - F_{T}= 0. For the box on the table we can write Sum(F) = ma: x-component: F_{T}- F_{fr}= 0; y-component: F_{N}- m_{1}g = 0. When we combine the equations, we have F_{fr}= F_{T}= m_{2}g = u_{s}F_{N}= u_{s}m_{1}g. Thus we have m_{1}- m_{2}/u_{s}= (2.0 kg)/(0.40) = 5.0 kg. (b) The acceleration is zero, so the only change is that the friction is kinetic: m_{1}- m_{2}/u_{k}= (2.0 kg)/(0.30) = 6.7 kg.

32. (a) The two blocks must have the same acceleration. From the force diagram for the top block we have x-component: F_{fr1}= M_{1}a; y-component: F_{N1}- M_{1}g = 0, or F_{N1}= M_{1}g. For the static friction force we have F_{fr1}= M_{1}a = u_{s}F_{N1}= u_{s}M_{1}g. Thus we have u_{s}= a/g = (5.2 m/s^{2})/(9.80 m/s^{2}) = 0.53. (b) If the coefficient of friction is less, the top block will slide. Because the friction force is one-half the force in part (a), the acceleration is also reduced by half: a_{1}= 1/2a = 1/2(5.2 m/s^{2}) = 2.6 m/s^{2}. (c) For the relative acceleration we have a_{12}= a_{1}- a = 2.6 m/s^{2}- 5.2 m/s^{2}= - 2.6 m/s^{2}. (d) From the force diagram for the bottom block we have x-component: F - F_{fr1}= M_{2}a; y-component: F_{N2}- F_{N1}- M_{2}g = 0, or F_{N2}= M_{1}g + M_{2}g. For part (a), if we add the two x-equations, we get F_{a}= M_{1}a + M_{2}a = (4.0 kg + 12.0 kg)(5.2 m/s^{2}) = 83 N. For part (b), if we add the two x-equations, we get F_{b}= M_{1}a1 + M_{2}a = (4.0 kg)(2.6 m/s^{2}) + (12.0 kg)(5.2 m/s^{2}) = 73 N.

44. The net force on Tarzan will provide his centripetal acceleration, which we take as the positive direction. We write Sum(F) = ma from the force diagram for Tarzan: F_{T}- mg = ma = mv^{2}/R. The maximum speed will require the maximum tension that Tarzan can create: 1400 N - (80 kg)(9.80 m/s^{2}) = (80 kg)v^{2}/(4.8 m), which gives v = 6.1 m/s.

74. We write Sum(F) = ma from the force diagram for the stationary hanging mass, with down positive: mg - F_{T}= ma = 0; which gives F_{T}= mg. For the rotating puck, the tension provides the centripetal acceleration, Sum(F_{R}) = Ma_{R}: F_{T}= Mv^{2}/R. When we combine the two equations, we have Mv^{2}/R = mg, which gives v = (mgR/M)^{1/2}.

92. (a) If the bead does not move along the hoop, it must have only the radial acceleration moving in a circle of radius R = r sinq. Since f = 1/t, then v = 2 pi R f = 2 pi r sinq f. We write Sum(F) = ma from the force diagram: x-component: F_{N}sinq = ma_{R}= mv^{2}/R; F_{N}sinq = mR(2pi f)^{2}= 4pi^{2}f^{2}mr sinq; y-component: F_{N}cosq - mg = 0, or F_{N}= mg /cosq. Combining these, we get cosq = g/4pi^{2}f^{2}r. (b) For the given data we get cosq = (9.80 m/s^{2})/4pi^{2}(4.0 rev/s)^{2}(0.20 m) = 0.0776, q = 86^{o}. (c) Because there is no friction, the bead cannot ride as high as the center. There would be no force to balance the downward force of gravity.

33. If the triangular block is pushed so that the small block tends to move up the incline, the static friction force on the small block will be down the incline, as shown. We choose a coordinate system with the x-axis in the direction of the acceleration a of the triangular block. From the force diagram for the small block we have x-component: F_{N}sinq + F_{fr}cosq = ma_{x}; y-component: F_{N}cosq - mg - F_{fr}sinq = ma_{y}. The top block will not slide until F_{fr}> mF_{N}. As long as this is not true, a_{x}= a, and a_{y}= 0. Thus we find the limiting acceleration by using these conditions: F_{N}cosq - mg - mF_{N}sinq = 0, or F_{N}= mg/(cosq - m sinq); F_{N}sinq + mF_{N}cosq = ma_{max}, or a_{max}= F_{N}(sinq + m cosq)/m = g(sinq + m cosq)/(cosq - m sinq). From the force diagram for the triangular block we have x-component: F - F_{N}sinq - F_{fr}cosq = Ma. Thus the maximum force F for the small block to not slide, and thus the minimum force to make the small block slide, is Fmin = F_{N}sinq + F_{fr}cosq + Ma_{max}= (m + M)a_{max}= (m + M)g(sinq + m cosq)/(cosq - u_{s}sinq).