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Figure 1, two people of the same heigt are holding up a box attached to a pole. The...

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kibauu | eNoter

Posted May 28, 2013 at 6:38 AM via web

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Figure 1, two people of the same heigt are holding up a box attached to a pole. The center of gravity  is located at point G.

Figure 2, the two people are now standing on a slope inclined by angle α(<phi/4) from horizontal. The person support at A and B exerts a force of magnitude Fa and Fb vertically upward. What is the value of ratio Fa/Fb?

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pramodpandey | College Teacher | Valedictorian

Posted May 28, 2013 at 7:47 AM (Answer #1)

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Force Fa is acting at A which is `(l cos(theta)+lsin(theta))` horizontal distance from shifted centre of gravity.

Force Fb is acting at B which is `(lcos(theta)-lsin(theta))` horizontal distance from shifted centre of gravity.

Thus

`l F_a (cos(theta)+sin(theta))=lF_b(cos(theta)-sin(theta))`

`(F_a)/(F_b)=(cos(theta)-sin(theta))/(cos(theta)+sin(theta))`

`=(1-tan(theta))/(1+tan(theta))`

`=tan(pi/4-theta)`

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saidctbb | Student, College Freshman | eNotes Newbie

Posted June 20, 2013 at 7:25 PM (Reply #1)

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i didn't understand why  Fa=l(cos(teta)+sin(teta))

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saidctbb | Student, College Freshman | eNotes Newbie

Posted June 20, 2013 at 7:24 PM (Answer #2)

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i didn't understand why  Fa=l(cos(teta)+sin(teta))

 

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