5. We choose the coordinate system shown, with positive torques clockwise. We write St = I a about the point A from the force diagram for the leg: St_{A}= MgD - F_{T}L = 0; (15.0 kg)(9.80 m/s^{2})(0.350 m) - F_{T}(0.805 m), which gives F_{T}= 63.9 N. Because there is no acceleration of the hanging mass, we have F_{T}= mg, or m = F_{T}/g = (63.9 N)/(9.80 m/s^{2}) = 6.52 kg.

13. We choose the coordinate system shown, with positive torques clockwise. We write St = I a about the support point A from the force diagram for the cantilever: St_{A}= - F_{2}d + Mg1/2 (d + D) = 0; - F_{2}(20.0 m) + (1200 kg)(9.80 m/s^{2}) 1/2(20.0 m + 30.0 m) = 0, which gives F_{2}= 1.47 x 10^{4}N. For the forces in the y-direction we have SF_{y}= F_{1}+ F_{2}- Mg = 0, or F_{1}= Mg - F_{2}= (1200 kg)(9.80 m/s^{2}) - (1.47 x 10^{4}N) = - 2.94 x 10^{3}N (down).

19. Because the person is standing on one foot, the normal force on the ball of the foot must support the weight: FN = Mg. We choose the coordinate system shown, with positive torques clockwise. We write St = I a about the point A from the force diagram for the bone in the foot: St_{A}= F_{T}d - F_{N}D = 0; F_{T}d - F_{N}(2d) = 0, which gives F_{T}= 2F_{N}= 2(70 kg)(9.80 m/s^{2}) = 1.4 x 10^{3}N (up). We write SF_{y}= ma_{y}from the force diagram: F_{T}+ F_{N}- F_{bone}= 0, which gives F_{bone}= F_{T}+ F_{N}= 3F_{N}= 3(70 kg)(9.80 m/s^{2}) = 2.1 x 10^{3}N (down).

27. Because the backpack is at the midpoint of the rope, the angles are equal. The force exerted by the backpacker is the tension in the rope. From the force diagram for the backpack and junction we can write SF_{x}= F_{T1}cosq - F_{T2}cosq = 0, or F_{T1}= F_{T2}= F; SF_{y}= F_{T1}sinq + F_{T2}sinq - mg = 0, or 2F sinq = mg. (a) We find the angle from tanq = h/(1/2)L = (1.5 m)/(1/2)(7.6 m) = 0.395, or q = 21.5°. When we put this in the force equation, we get 2F sin 21.5° = (16 kg)(9.80 m/s^{2}), which gives F = 2.1 x 10^{2}N. (b) We find the angle from tanq = h/(1/2)L = (0.15 m)/(1/2)(7.6 m) = 0.0395, or q = 2.26°. When we put this in the force equation, we get 2F sin 2.26° = (16 kg)(9.80 m/s^{2}), which gives F = 2.0 x 10^{3}N.

39. From the symmetry of the wires, we see that the angle between a horizontal line on the ground parallel to the net and the line from the base of the pole to the anchoring point is q = 30°. We find the angle between the pole and a wire from tanb = d/H = (2.0 m)/(2.6 m) = 0.769, which gives b = 37.6°. Thus the horizontal component of each tension is F_{T}sinb. We write St = I a about the horizontal axis through the base A perpendicular to the net from the force diagram for the pole: St_{A}= F_{net}H - 2(F_{T}sinb cosq)H = 0, or F_{net}= 2F_{T}sinb cosq = 2(95 N) sin 37.6° cos 30° = 1.0 x 10^{2}N.

47. The pressure needed is determined by the bulk modulus: DP = - B DV/V_{0}= - (90 x 10^{9}N/m^{2})(- 0.10 x 10^{-2}) = 9.0 x 10^{7}N/m^{2}. This is (9.0 x 10^{7}N/m^{2})/(1.0 x 10^{5}N/m^{2}· atm) = 9.0 x 10^{2}atm.

12. From the force diagram for the hanging light and junction we can write SF_{x}= F_{T1}cosq_{1}- F_{T2}cosq_{2}= 0; F_{T1}cos 37° = F_{T2}cos 53°; SF_{y}= F_{T1}sinq_{1}+ F_{T2}sinq_{2}- mg = 0; F_{T1}sin 37° + F_{T2}sin 53° = (30 kg)(9.80 m/s^{2}). When we solve these two equations for the two unknowns, F_{T1}, and F_{T2}, we get F_{T1}= 1.8 x 10^{2}N, and F_{T2}= 2.3 x 10^{2}N.

16. Because F_{T}= F_{B}, from the symmetry we see that F_{1}= F_{2}. We choose one handle as the system and write St=Ia about the hinge P: St_{P}= F_{T}L_{1}cosq - F_{2}L_{2}cosq = 0, which gives F_{2}= F_{T}L_{1}/L_{2}= (11.0 N)(8.50 cm)/(2.70 cm) = 34.6 N.

26. We choose the coordinate system shown, with positive torques clockwise. We write St=Iaabout the point A from the force diagram for the beam: St_{A}= - (F_{T}sinq)L + Mg1/2L = 0; - F_{T}sin 40° + (30 kg)(9.80 m/s^{2})1/2 = 0, which gives F_{T}= 2.3 x 10^{2}N. Note that we find the torque produced by the tension by finding the torques produced by the components. We write SF = ma from the force diagram for the beam: SF_{x}= F_{Wx}- F_{T}cosq = 0; F_{Wx}- (2.29 x 10^{2}N) cos 40° = 0, which gives F_{Wx}= 175 N. SF_{y}= F_{Wy}+ F_{T}sinq - Mg = 0; F_{Wy}+ (2.29 x 10^{2}N) sin 40° - (30 kg)(9.80 m/s^{2}) = 0, which gives F_{Wy}= 147 N. For the magnitude of F_{W}we have F_{W}= (F_{Wx}2 + F_{Wy}2)1/2 = [(175 N)2 + (147 N)2]1/2 = 2.3 x 10^{2}N. We find the direction from tanq = F_{Wy}/F_{Wx}= (147 N)/(175 N) = 0.84, which gives q= 40°. Thus the force at the wall is F_{W}= 2.3 x 10^{2}N, 40° above the horizontal.

32. We choose the coordinate system shown, with positive torques counterclockwise. We write St=Ia about the point A from the force diagram for the ladder: St_{A}= mg(1/2¬) cosq - F_{N2}r sinq = 0, which gives F_{N2}= mg/2 tanq. We write SF_{x}= ma_{x}from the force diagram for the ladder: F_{fr}- F_{N2}= 0, which gives F_{fr}= F_{N2}= mg/2 tanq. We write SF_{y}= ma_{y}from the force diagram for the ladder: F_{N1}- mg = 0, which gives F_{N1}= mg. For the bottom not to slip, we must have F_{fr}<= m_{s}F_{N1}, or mg/2 tanq ? ?smg, from which we get tanq >= (1/2 m_{s}). The minimum angle is q_{min}= tan^{-1}(1/2 m_{s}).

52. We find the maximum compressive force from the compressive strength of bone: F_{max}= (Compressive strength) Area = (170 x 10^{6}N/m^{2})(3.0 x 10^{-4}m^{2}) = 5.1 x 10^{4}N.

69. All elements are in equilibrium. For the C-D pair, we write St=Iaabout the point c from the force diagram: St_{c}= M_{C}gL_{6}- M_{D}gL_{5}= 0; M_{C}(5.00 cm) - M_{D}(17.50 cm) = 0, or M_{C}= 3.500M_{D}. The center of mass of C and D must be under the point c. We write St=Iaabout the point b from the force diagram: St_{b}= M_{B}gL_{4}- (M_{C}+ M_{D})gL_{3}= 0; (0.735 kg)(5.00 cm) - (M_{C}+ M_{D})(15.00 cm) = 0, or M_{C}+ M_{D}= 0.245 kg. When we combine these two results, we get M_{C}= 0.191 kg, and M_{D}= 0.0544 kg. The center of mass of B, C, and D must be under the point b. For the entire mobile, we write St=Ia about the point a from the force diagram: St_{a}= (M_{B}+ M_{C}+ M_{D})gL_{2}- M_{A}gL_{1}= 0; (0.735 kg + 0.245 kg)(7.50 cm) - M_{A}(30.00 cm) = 0, which gives M_{A}= 0.245 kg.

71. (a) The cylinder will roll about the contact point A. We write St = I_{A}a a about the point A: F_{a}(2R - h) + F_{N1}[R^{2}- (R - h)^{2}]^{1/2}- Mg[R^{2}- (R - h)^{2}]^{1/2}= I_{A}a. When the cylinder does roll over the curb, contact with the ground is lost and F_{N1}= 0. Thus we get F_{a}= {I_{A}a + Mg[R^{2}- (R - h)^{2}]^{1/2}}/(2R - h) = [I_{A}a/(2R - h)] + [Mg(2Rh - h^{2})^{1/2}/(2R - h)]. The minimum force occurs when a = 0: F_{amin}= Mg[h(2R - h)]^{1/2}/(2R - h) = Mg[h/(2R - h)]^{1/2}. (b) The cylinder will roll about the contact point A. We write St = I_{A}a about the point A: F_{b}(R - h) + F_{N1}[R^{2}- (R - h)^{2}]^{1/2}- Mg[R^{2}- (R - h)^{2}]^{1/2}= I_{A}a. When the cylinder does roll over the curb, contact with the ground is lost and F_{N1}= 0. Thus we get F_{b}= {I_{A}a + Mg[R^{2}- (R - h)^{2}]^{1/2}}/(R - h) = [I_{A}a/(R - h)] + [Mg(2Rh - h^{2})^{1/2}/(R - h)]. The minimum force occurs when a = 0: F_{bmin}= Mg[h(2R - h)]^{1/2}/(R - h).