Problem The force P has a magnitude

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Presentation transcript:

Problem 3.154 The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming a = 30o and b = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. a b P C B A 200 mm 300 mm

Solving Problems on Your Own a b P C B A 200 mm 300 mm The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming a = 30o and b = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. 1. Replace a force with an equivalent force-couple system at a specified point. The force of the force-couple system is equal to the original force, while the required couple vector is equal to the moment of the original force about the given point.

S F : F = P or F = 250 N 60o S MB : M = _ (0.3 m)(250 N) = _75 N Problem 3.154 Solution a b P C B A 200 mm 300 mm Replace a force with an equivalent force-couple system at a specified point. (a) Equivalence requires: S F : F = P or F = 250 N 60o S MB : M = _ (0.3 m)(250 N) = _75 N The equivalent force couple system at B is: F = 250 N 60o, M = 75 N . m

(b) Require: The two force systems are equivalent. Problem 3.154 Solution a b P C B A 200 mm 300 mm (b) Require: The two force systems are equivalent. B A C 60o 250 N 30o F FB FA x y =

S Fy : _250 = _ FA sin F _ FB sin F If FA = _ FB then _ 250 = 0 reject Problem 3.154 Solution B A C 60o 250 N 30o F FB FA x y = Equivalence then requires: S Fx : 0 = FA cos F + FB cos F FA = _ FB or cos F = 0 S Fy : _250 = _ FA sin F _ FB sin F If FA = _ FB then _ 250 = 0 reject Consequently cos F = 0 or F = 90o and FA + FB = 250

S MB : _ (0.3 m)( 250 N) = (0.2 m) FA or FA = _ 375 N and FB = + 675 N Problem 3.154 Solution B A C 60o 250 N 30o x y C B = F A FB F FA Also: + S MB : _ (0.3 m)( 250 N) = (0.2 m) FA or FA = _ 375 N and FB = + 675 N FA = 375 N 60o, FB = 625 N 60o