EX17.1-Force on Hanging Bar

# EX17.1-Force on Hanging Bar - n O G t O G G O G O G G Gn Gt...

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6/43. A bar is pinned at point O which weighs 10 lbs. Gages record a peak horizontal force = 48 lbs at point O, what is peak value of P? Kinetics so use Newton’s second law. Draw FBD, MAD t n You could also use x and y. Lets use equations of motion in n and t. O n – mg = m(L/2) ϖ 2 P + O t = ma Gt O n = mg = 10 lbs up We don’t know P or a - only have one equation. Do a moment equation. Let’s choose G as our point – sum moments about G. Now don’t know P, a Gt , or α , and have 2 equations (the n equation doesn’t really help us much here). Can we come up with another equation? Kinematics! We have fixed axis rotation so we can use the vector equation:
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Unformatted text preview: n O G t O G G O G O G G Gn Gt G e r e r a r r a a a a O O O O O O O O O O O 2 / / / / ϖ α + = × × + × = + = We know that r G/O = L/2 and ω=0 (no angular velocity) Sub in: a Gt =L/2α o Could also do about O Solution: Problem Type: Find: Given: l=18”, mg=10 lbs, O x = 48 lbs Value of P Rigid Body Kinetics, Fixed Axis Rotation EX17.1 – Force on a Hanging Bar ∑F n = ma n G (2) Reflection: Kinematics in Rigid Body motion can give us equations to relate the acceleration of the mass center to the angular acceleration of the body. α=618.24 rad/s 2 (1) _ P=96 lbs Sketch:...
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## This note was uploaded on 01/22/2011 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.

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