JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year ENGINEERING MECHANICS Previous Question Paper (March - 2021 - R18)

To find the resultant of a coplanar force system, where all forces lie in the same plane, the following analytical procedure is used:

• Resolution of Forces: Resolve each individual force F_i into its rectangular components along the mutually perpendicular x and y axes. If a force makes an angle θ_i with the positive x-axis, the components are:
  F_ix = F_i cos(θ_i)
  F_iy = F_i sin(θ_i)
• Summation of Components: Calculate the algebraic sum of all x-components and y-components separately:
  ΣF_x = Σ F_i cos(θ_i)
  ΣF_y = Σ F_i sin(θ_i)
• Magnitude of Resultant: The magnitude of the resultant force R is determined using the Pythagorean theorem:
  R = √(ΣF_x² + ΣF_y²)
• Direction of Resultant: The angle α that the resultant makes with the x-axis is given by:
  α = tan^-1(|ΣF_y / ΣF_x|)

In a regular hexagon, the interior angles are 60° apart. Let the forces act from the center towards the vertices at angles 0°, 60°, 120°, 180°, 240°, and 300°.

• Forces: F₁=15, F₂=25, F₃=35, F₄=45, F₅=55, F₆=65 (all in kN).
• Sum of x-components (ΣF_x):
  ΣF_x = 15cos(0°) + 25cos(60°) + 35cos(120°) + 45cos(180°) + 55cos(240°) + 65cos(300°)
  ΣF_x = 15(1) + 25(0.5) + 35(-0.5) + 45(-1) + 55(-0.5) + 65(0.5) = 15 + 12.5 - 17.5 - 45 - 27.5 + 32.5 = -30 kN.
• Sum of y-components (ΣF_y):
  ΣF_y = 15sin(0°) + 25sin(60°) + 35sin(120°) + 45sin(180°) + 55sin(240°) + 65sin(300°)
  ΣF_y = 15(0) + 25(√3/2) + 35(√3/2) + 45(0) + 55(-√3/2) + 65(-√3/2)
  ΣF_y = (√3/2) * (25 + 35 - 55 - 65) = (√3/2) * (-60) = -30√3 ≈ -51.96 kN.
• Resultant Magnitude (R):
  
  R = √((-30)² + (-51.96)²) = √(900 + 2700) = √3600 = 60 kN
• Direction (α):
  
  α = tan^-1(51.96 / 30) = 60°
  
  Since both components are negative, the resultant lies in the 3rd quadrant at 180° + 60° = 240° from the positive x-axis.