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Unformatted text preview: 1. Find the magnitude of the resultant of the two forces in Figure 1 and its direction measured counterclockwise from the xdirection. 250 N 400 N x y 60 o 45 o Figure 1 Solution The 250 N force can be expressed as the Cartesian vector { 250 sin 60 , 250 cos 60 } = { 216 . 5 , 125 } N and the 400 N force as { 400 sin 45 , 400 cos 45 } = { 282 . 8 , 282 . 8 } N . Thus the resultant is { 216 . 5 282 . 8 , 125 + 282 . 8 } = { 66 . 3 , 407 . 8 } N with magnitude √ 66 . 3 2 + 407 . 8 2 = 413 . 2 N . The counterclockwise angle from the xaxis is θ where cos θ = 66 . 3 413 . 2 or θ = 99 . 23 o . 1 2. Two cables with known tensions 20 kN and 45 kN are attached to the top of the plon AB in Figure 2. A third cable BC is used as a guy wire. Find the tension in BC , knowing that the resultant force applied to the end B of the pylon by the three cables must be vertical. 15 o 25 o 20 kN 45 kN A B C 32 m 14 m Figure 2 Solution If the resultant force exerted on the top of the pylon at B is vertical, its horizontal component must be zero. This horizontal component can be written as 20 cos 25 45 cos 15 + 14 T √ 14 2 + 32 2 kN , where T is the tension in the cable BC and we have used the fact that the rightangle triangle ABC has sides 14 and 32 meters to resolve this force horizontally. Setting the horizontal component to zero, we have 18 . 126 43 . 467 + 0 . 4008 T = 0 or T = 43 . 467 18 . 126 . 4008 = 63 . 23 kN . 2 3. The hinged plate in Figure 3 is supported by the cord AB . If the tension in the cord is 340 lb, express the force F that it exerts on the plate as a Cartesian vector. Also, find the length of the cord. A B x y z 8 9 6 all dimensions in feet Figure 3 Solution The points A,B are defined by the position vectors r A = { 9 , , 8 } r B = { , 6 , } , so r AB = r B r A = { 9 , 6 , 8 } = { 9 , 6 , 8 } feet and the length of the cord is  r AB  = √ 9 2 + 6 2 + 8 2 = 13 . 45 feet ....
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 Fall '07
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 Force, Cartesian vector, position vectors

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