Sample Problem: Solution [PDF]

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1.4



Sample Problem



Figure (a) shows two position vectors of magnitudes A = 60 ft and B = 100 ft. (A position vector is a vector drawn between two points in space.) Determine the resultant R = A + B using the following methods: (1) analytically, using the triangle law; and (2) graphically, using the triangle law.



100



ft



B



A 60



ft



70°



30° (a)



Solution Part 1



B=



100



ft



The first step in the analytical solution is to draw a sketch (approximately to scale) of the triangle law. The magnitude and direction of the resultant are then found by applying the laws of sines and cosines to the triangle. In this problem, the triangle law for the vector addition of A and B is shown in Fig. (b). The magnitude R of the resultant and the angle α are the unknowns to be determined. Applying the law of cosines, we obtain



R 140°



α



A=



60



30° (b)



70°



R 2 = 602 + 1002 − 2(60)(100) cos 140◦ which yields R = 151.0 ft. The angle α can now be found from the law of sines:



ft



R 100 = sin α sin 140◦ Substituting R = 151.0 ft and solving for α, we get α = 25.2◦ . Referring to Fig. (b), we see that the angle that R makes with the horizontal is 30◦ + α = 30◦ + 25.2◦ = 55.2◦ . Therefore, the resultant of A and B is R = 151.0 ft



Answer 55.2°



4



Part 2 In the graphical solution, Fig. (b) is drawn to scale with the aid of a ruler and a protractor. We first draw the vector A at 30◦ to the horizontal and then append vector B at 70◦ to the horizontal. The resultant R is then obtained by drawing a line from the tail of A to the head of B. The magnitude of R and the angle it makes with the horizontal can now be measured directly from the figure. Of course, the results would not be as accurate as those obtained in the analytical solution. If care is taken in making the drawing, two-digit accuracy is the best we can hope for. In this problem we should get R ≈ 150 ft, inclined at 55◦ to the horizontal.



1.5



Sample Problem



The vertical force P of magnitude 100 kN is applied to the frame shown in Fig. (a). Resolve P into components that are parallel to the members AB and AC of the truss.



A



B



P



P = 100 kN



70° 35° PAC 110°



35° 70° C



35° PAB



(a)



(b)



Solution The force triangle in Fig. (b) represents the vector addition P = P AC + PAB . The angles in the figure were derived from the inclinations of AC and AB with the vertical: P AC is inclined at 35◦ (parallel to AC), and PAB is inclined at 70◦ (parallel to AB). Applying the law of sines to the triangle, we obtain



100 PAB PAC = = sin 35◦ sin 35◦ sin 110◦



which yields for the magnitudes of the components



PAB = 100.0 kN



PBC = 163.8 kN



Answer



5



Problems Set



2



Problems v1



30° 60°



v2



1.9 The magnitudes of the two velocity vectors are v1 = 3 m/s and v2 = 2 m/s. Determine their resultant v = v1 + v2 . 1.10 Determine the magnitudes of vectors v1 and v2 so that their resultant is a



Fig. P1.9, P1.10



horizontal vector of magnitude 4 m/s directed to the right.



y 200 lb



1.11 Resolve the 200-lb force into components along (a) the x- and y-axes and (b) the x  - and y-axes.



30° x 20° x'



Fig. P1.11



1.12



The 500-N weight is supported by two cables, the cable forces being F1 and F2 . Knowing that the resultant of F1 and F2 is a force of magnitude 500 N acting in the y-direction, determine F1 and F2 . y F2



F1 50°



35°



500 N



Fig. P1.12



1.13 Resolve the 360-lb force into components along the cables AB and AC. Use α = 55◦ and β = 30◦ .



1.14 The supporting cables AB and AC are oriented so that the components of the 360-lb force along AB and AC are 185 lb and 200 lb, respectively. Determine the angles α and β. B



C



α



β



A 360 lb



Fig. P1.13, P1.14