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D C B A aaa  30 o P Problem 4-e Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing.

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Presentation on theme: "D C B A aaa  30 o P Problem 4-e Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing."— Presentation transcript:

1 D C B A aaa  30 o P Problem 4-e Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle  = 30 o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C.

2 D C B A aaa  30 o P 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. Solving Problems on Your Own Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle  = 30 o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. Problem 4-e

3 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be:  F x = 0  F y = 0  M O = 0 where O is an arbitrary point in the plane of the structure or  F x = 0  M A = 0  M B = 0 where point B is such that line AB is not parallel to the y axis or  M A = 0  M B = 0  M C = 0 where the points A, B, and C do not lie in a straight line. D C B A aaa  30 o P Solving Problems on Your Own Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle  = 30 o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. Problem 4-e

4 Problem 4-e Solution Draw a free-body diagram of the body. D C B A aaa  30 o P D C BA P aaa C B T

5 Write equilibrium equations and solve for the unknowns. Problem 4-e Solution D C BA 30 o P aaa C B T  F = 0: _ P cos 30 o + T cos 60 o = 0 T = P = P T = 3 P 30 o cos 30 o cos 60 o 3 / 2 1 / 2 +  M B = 0: P a _ (C sin 30 o ) a + T cos 30 o (2a) = 0 P a _ ( C ) a + 3 P ( ) 2a = 0 _ C + (1 + 3) P = 0; C = 8 P C = 8 P 30 o 1 2 3 2 1 2

6 D C BA 30 o P aaa C B T  +  F = 0: _ B cos 30 o + C cos 30 o _ T sin 30 o = 0 _ B + 8 P _ 3 P ( ) = 0; B = 7 P B = 7 P 30 o 3 2 3 2 1 2 Problem 4-e Solution


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