**Unformatted text preview: **) ( ) ( 2 1 2 2 2 x x d k gd M m v M m-+ = + + + , ) (( 2 1 ) ( ) 2 ( 2 1 2 2 2 x x d x mg gd M m gh M m m-+ = + + + where the above expression for v , and x mg k = have been used. In this form, it is seen that a factor of g cancels from all terms. After performing the algebra, the quadratic for d becomes which has as its positive root For this situation, m = 4/3 M and h/x 0 = 6, so d = 0.232 m. , 2 2 2 2 = +- -M m m hx M m x d d . ) ( 2 2 2 + + + = M m M m x h M m M m x d...

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