This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Okay, 🤍🦋Mashiyamahle🤍🦋, let's continue with these problems. Question 5: Step 1: Convert the diameter to meters and the force to Newtons. The diameter d = 12 mm = 0.012 m. The radius r = (d)/(2) = 0.012 m2 = 0.006 m. The force F = 80 kN = 80 × 10^3 N. Step 2: Calculate the cross-sectional area of the bolt. Assuming single shear for two sheets bolted together: A = r^2 = 3.142 × (0.006 m)^2 = 3.142 × 0.000036 m^2 = 0.000113112 m^2 Step 3: Calculate the maximum shear stress. = (F)/(A) = 80 × 10^3 N0.000113112 m^2 = 707259000 Pa Converting to MPa: = 707.26 MPa. The maximum shear stress for the bolt is 707.26 MPa Question 6: a) The normal stress due to the axial force. Step 1: Convert dimensions to meters and the axial force to Newtons. The width w = 80 cm = 0.80 m. The height h = 20 cm = 0.20 m. The axial force F_axial = 12 kN = 12 × 10^3 N. Step 2: Calculate the cross-sectional area. A = w × h = 0.80 m × 0.20 m = 0.16 m^2 Step 3: Calculate the normal stress. = F_axialA = 12 × 10^3 N0.16 m^2 = 75000 Pa Converting to kPa: = 75 kPa. The normal stress is 75 kPa b) The shear stress due to a shear force of 18 kN. Step 1: Convert the shear force to Newtons. The shear force F_shear = 18 kN = 18 × 10^3 N. The cross-sectional area is A = 0.16 m^2 (from part a). Step 2: Calculate the average shear stress. = F_shearA = 18 × 10^3 N0.16 m^2 = 112500 Pa Converting to kPa: = 112.5 kPa. The shear stress is 112.5 kPa Question 7: Step 1: Calculate the total weight of the bar. The mass m = 500 kg. Assume acceleration due to gravity g = 9.81 m/s^2. The total weight W = m × g = 500 kg × 9.81 m/s^2 = 4905 N. Step 2: Determine the force supported by each cable. Since the bar is homogeneous and supported at both ends by two cables, the weight is evenly distributed. The force per cable F_cable = (W)/(2) = 4905 N2 = 2452.5 N. Step 3: Calculate the smallest area for a copper cable. The maximum allowable stress for copper _Cu = 80 MPa = 80 × 10^6 Pa. A_Cu = F_cable_Cu = 2452.5 N80 × 10^6 Pa = 0.00003065625 m^2 Converting to mm^2: A_Cu = 0.00003065625 m^2 × (1000 mm/m)^2 = 30.65625 mm^2. The smallest area for a copper cable is 30.66 mm^2 Step 4: Calculate the smallest area for a steel cable. The maximum allowable stress for steel _St = 100 MPa = 100 × 10^6 Pa. A_St = F_cable_St = 2452.5 N100 × 10^6 Pa = 0.000024525 m^2 Converting to mm^2: A_St = 0.000024525 m^2 × (1000 mm/m)^2 = 24.525 mm^2. The smallest area for a steel cable is 24.53 mm^2 Question 8: Step 1: Convert the diameter and thickness to meters, and shear strength to Pascals. The hole diameter d = 15 mm = 0.015 m. The plate thickness t = 20 mm = 0.020 m. The shear strength = 300 MN/m^2 = 300 × 10^6 Pa. Step 2: Calculate the shear area. A = d t = 3.142 × 0.015 m × 0.020 m = 0.0009426 m^2 Step 3: Calculate the force required. F = × A = 300 × 10^6 Pa × 0.0009426 m^2 = 282780 N Converting to kN: F = 282.78 kN. The force required is 282.78 kN Question 9: Step 1: Convert the force to Newtons and the shearing strength to Pascals. The force F = 380 kN = 380 × 10^3 N. The shearing strength = 400 MPa = 400 × 10^6 Pa. Step 2: Calculate the required total shear area. A_total = (F)/() = 380 × 10^3 N400 × 10^6 Pa = 0.00095 m^2 Step 3: Calculate the area of one shear plane for a clevis joint (double shear). For a clevis, the force is resisted by two shear planes, so A_total = 2 × A_pin. A_pin = A_total2 = 0.00095 m^22 = 0.000475 m^2 Step 4: Calculate the diameter of the bolt. The area of one pin is A_pin = ((d)/(2))^2. 0.000475 m^2 = 3.142 × ((d)/(2))^2 ((d)/(2))^2 = 0.000475 m^23.142 = 0.00015118 m^2 (d)/(2) = sqrt(0.00015118 m)^2 = 0.01230 m d = 2 × 0.01230 m = 0.02460 m Converting to millimeters: d = 0.02460 m × 1000 mm/m = 24.60 mm. The smallest diameter bolt is 24.60 mm That's 2 down. 3 left today — send the next one.