Problem 4-b 4 in 4 in 20 lb 40 lb The T-shaped bracket shown is

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Equilibrium of Rigid Bodies

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Presentation transcript:

Problem 4-b 4 in 4 in 20 lb 40 lb The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. A B 2 in C D 3 in 3 in E q

4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. Problem 4-b 20 lb 40 lb A B 2 in C D 3 in 3 in E q 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. Problem 4-b 20 lb 40 lb A B 2 in C D 3 in 3 in E q 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

Draw a free-body diagram of the body. 2 in C D 3 in 4 in 4 in 3 in E Problem 4-b Solution 20 lb 40 lb A B Draw a free-body diagram of the body. 2 in C D 3 in 4 in 4 in 3 in E 20 lb 40 lb q A B C 2 in C D 3 in D 3 in E E 30o

+ S Fy = 0: E cos 30o _ 20 _ 40 = 0 E = = 69.28 lb 60 lb 4 in 4 in Problem 4-b Solution 20 lb 40 lb A B C 2 in Write equilibrium equations and solve for the unknowns. C D 3 in D 3 in E E 30o + S Fy = 0: E cos 30o _ 20 _ 40 = 0 E = = 69.28 lb E = 69.3 lb 60o 60 lb cos 30o

( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0 Problem 4-b Solution 20 lb 40 lb A B C 2 in C D 3 in D 3 in E E + S MD = 0: ( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0 _ 80 _ 3C + 69.28 ( 0.5 )( 3 ) = 0 C = 7.974 lb C = 7.97 lb 30o

4 in 4 in Problem 4-b Solution 20 lb 40 lb A B C 2 in C + S Fx= 0: E sin 30o + C _ D = 0 ( 69.28 lb )( 0.5 ) + 7.974 lb _ D = 0 D = 42.6 lb D 3 in D 3 in E E 30o