Tiêu đề C1 PSAD Reviewer.pdf ✅

C1 PSAD Reviewer ^_^ 4. If the resultant of the forces is 80 kN and is acting along the positive x-axis, find angle α. A. 60° C. 22.85° B. 35° D. 14.78° 5. If α = 60°, what is the value of force C such that the resultant of forces A, B, and C acts along the x-axis? A. 45 kN C. 35 kN B. 80 kN D. 40 kN 6. Find the value of C for A, B, and C to be in equilibrium. A. 45 kN C. 72.17 kN B. 40.9 kN D. 36.41 KN Solution: Difficulty: 1/5 C cosα + 45cos60° - 35 = 80 C cosα = 92.5 C sinα = 45sin60° C sinα = 38.97 tanα = 38.97 92.5 ; α = 22.85° If α = 60° C sin60° = 45sin60° ; C = 45kN C cosα + 45cos60° = 35 C cosα = 12.5 C sinα = 45sin60° C sin= 338.97
C1 PSAD Reviewer ^_^ tanα = 38.97 12.5 ; α = 72.2° C sin722° = 38.97 ; C = 40.9kN Situation 3 - A man supports himself and the uniform horizontal beam by pulling on the rope with force T. The weight of the man and the beam are 883 N and 245 N, respectively. 7. Compute the tension in the rope. A. 132.5 N C. 442 N B. 245 N D. 387.5 N 8. Compute the horizontal reaction at A. A. 442.5 N C. 387.5 N B. 132.5 N D. 785.12 N 9. Compute the vertical reaction at A. A. 376.4 N C. 358.21 N B. 245 N D. 452.14 N Solution: Difficulty: 1/5 ΣMa = 0 245(2.4)+883(3) + Tsin20°(2.8) = T(4.8)+Tcos20°(4.8) T = 387.5N ΣFh = 0 Ax = Tsin20° Ax = 387.5sin20° Ax = 132.5N ΣFv = 0 Ay + T + Tcos20° = 245+883 Ay + 387.5 + 387.5cos20° = 245+883
C1 PSAD Reviewer ^_^ Ay = 376.4 N Situation 4 - Load W is to be lifted using the crane which is hinged at B as shown in the figure. The value of x1 = 10m, x2 = 18m. Neglecting the weight of the crane, 10. Determine the force of Cable AC. A. 26.65 kN C. 42.85 kN B. 30.96 kN D. 19.74 kN 11. Determine the resultant reaction at B. A. 39.4 kN C. 37.48 kN B. 31.52 kN D. 25.96 kN 12. Determine the largest load that can be lifted if the maximum force of Cable AC is 50 kN. A. 67.84 kN C. 44.14 kN B. 47.84 kN D. 34.25 kN Solution: Difficulty: 1/5 tanΘ = 18 18 ; Θ = 45° tanα = 18 8 ; α = 66.04° ΣFh = 0 AC cos45° = BC sin23.96° BC = 1.74 AC ΣFv = 0 AC sin45° + 20 = BC cos23.96° 1.74(AC)cos23.96° = 20 + AC sin45° AC = 22.65 kN