Meriam, Kraige & Bolton — Statics of Particles: Problem 2_14

⚡ Mecademy AIENG정역학 · ch2  Problem Statement 2/14 The two forces shown act in the - plane of the T-beam cross section. If it is known that the resultant of the two forces has a magnitude of and a line of action that lies above the negative -axis, determine the magnitude of and the inclination of . Problem 2/14 ● 1. Coordinate System and Vector Setup Based on the problem diagram and standard conventions, we define the origin at the junction of the T-beam cross-section. The -axis is positive to the right, and the -axis is positive upward. Resultant : The magnitude is and it acts above the negative - axis (2nd quadrant). Force : Acting from the origin into the 2nd quadrant (pulling away), it makes an angle of with the positive -axis. Force : Acting from the 1st quadrant towards the origin (pushing), it makes an angle with the positive -axis. Its vector direction is therefore in the 3rd quadrant ( ). ● 2. Equilibrium Equations (Vector Summation) The resultant is the sum of the two forces: . We equate the scalar components: 1. Sum of -components: xy R3.5 kN15 ∘ xF 1 θF 2 O xy RR=3.5 kN15 ∘ x R=R(−cos15i+ ∘ sin15j)= ∘ 3.5(−0.9659i+0.2588j) R=−3.3807i+0.9059j [kN] F 1 30 ∘ y F = 1 −F sin30i+ 1 ∘ F cos30j= 1 ∘ −0.5F i+ 1 0.8660F j 1 F 2 θx180+ ∘ θ F = 2 2.8(−cosθi−sinθj) R=F + 1 F 2 x R = x F + 1x F ⟹ 2x −3.3807=−0.5F − 1 2.8cosθ 2. Sum of -components: ● 3. Solving the System of Equations Square both equations and add them to eliminate using the identity : Applying the quadratic for