MAE 101 homework 1 sol

# MAE 101 homework 1 sol - MAE 101 Statics and Strength of...

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MAE 101: Statics and Strength of Materials September 27, 2011 Problem 1 The unstretched length of the spring AB is 660 mm, and the spring constant k = 1000 N/m. What is the mass of the suspended object? 350 mm B 400 mm 600 mm A Solution: Use the linear spring force-extension relation to ﬁnd the magnitude of the tension in spring AB . Isolate juncture A . The forces are the weight and the tensions in the cables. The angles are tan α = ( 350 600 ) = 0 . 5833 , α = 30 . 26 (1) tan β = ( 350 400 ) = 0 . 875 , β = 41 . 2 (2) The angle between the x axis and the spring is α . The tension is T AB = | T AB | ( i cos α + j sin α ) . The angle between the x axis and AC is (180 - β ). The tension is T AC = | T AC | ( i cos(180 - β ) + j sin(180 - β )) T AC = | T AC | ( - i cos β + j sin β ) . The weight is: W = 0 i - | W | j . 1

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350 mm B 400 mm 600 mm A x y C A B ` _ W The equilibrium conditions: X F = W + T AB + T AC = 0 . Substitute and collect like terms X F x = ( | T AB | cos α - | T AC | cos β ) i = 0 , X F y = ( | T AC | sin α + | T AB | sin β - | W | ) j = 0 . Solve: | T AC | = ± cos α cos β ² | T AB | and | W | = ± sin 2 α +sin 2 β 2 cos β ² | T AB | .
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## This note was uploaded on 11/06/2011 for the course MAE 101 taught by Professor Orient during the Spring '08 term at UCLA.

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MAE 101 homework 1 sol - MAE 101 Statics and Strength of...

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