advertisement

Solving Non-right Triangle Vector Problems (using right triangle trig) Oh, Boy! Here’s the problem: Add the following displacements to get a resultant displacement: dA = 25.0m @ 33.0 dB = 42.0m @ 64.0 dres = ? Step 1: Sketch the vectors and the resultant dA + dB = dres Add them head to tail (start B from the head of A) Step 2: Resolve A & B into perpendicular components (x and y components) By Bx Ay Ax Notice that if you add Bx to Ax, they combine to make the base of a big right triangle whose hypotenuse is R! You can do the same with Ay and By to make the other side of the big right triangle. By Bx Ay Ax Ay Bx Step 3: Solve for Ax and Bx Ax = hyp (cosθ) Ax = 25.0m(cos33.0 ) Ax = 21.0 m By Bx = hyp (cosθ) Bx = 42.0m(cos64.0 ) Bx = 18.4 m 64.0 Bx Ay Ay 33.0 Ax= 21.0m Bx= 18.4m Step 4: Add Ax + Bx to make Rx—the base of the big right triangle Rx = Ax + Bx Rx = 21.0m + 18.4m Rx = 39.4 m By 64.0 Bx Ay 33.0 Ax Bx Rx = Ax + Bx = 39.4 m Ay Step 5: Solve for Ay and By Ay = hyp (sinθ) Ay = 25.0m(sin33.0 ) Ay = 13.6 m By = 37.7 m By = hyp (sinθ) By = 42.0m(sin64.0 ) By = 37.7 m 64.0 Bx Ay Ay = 13.6 m 33.0 Ax Bx Ry = Ay + By Ry = 13.6m + 37.7m Ry = 51.3 m By 64.0 Bx Ay Ay 33.0 Ax Bx Ry = Ay + By = 51.3 m Step 6: Add Ay + By to make Ry—the “height” of the big right triangle c2 = a2 + b2 R2 = Rx2 + Ry2 R2 = (39.4m)2 + (51.3m)2 R = 64.5 m Rx = Ax + Bx = 39.4 m Ry = Ay + By = 51.3 m Step 7: Solve for the hypotenuse of the big right triangle—the magnitude of the resultant displacement tan θ = opp./adj. tan θ = 51.3m / 39.4m θ = tan-1(51.3m / 39.4m) θ = 52.5° θ = 52.5° Rx = Ax + Bx = 39.4 m Ry = Ay + By = 51.3 m Step 8: Solve for the direction of the resultant vector by solving for angle θ within the big triangle. Since you have the lengths of all three sides now, you can use sine, cosine or tangent. Final answer? The resultant displacement dres = 64.5 m @ 52.5°