{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# pg 62 - 5.6—“ 1 Jﬁb 153m Our drawings for Example 4.5...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.6—“ 1 Jﬁb . 153m Our drawings for Example 4.5. es areih the-vertical'direeﬁqn, so we’re. Concerned laminate The fare with only the )2 Component of Newten’s. law: Fwy+ng=meyu1he spring force is upward‘and; fromHonkEklaw; it has magnitude kx, so I ' The )1 'c'emponent of New - which we solve to get mkx, 'Graxiity' is downward with magnimde mg, so F”: #1113. ‘ Fey ton’s law then becomes kx #mg = may, 111013, ‘1”- g) .. . 1: » Putting in- the numbers (a) wiih the helicopter at rest (ay :11) and (b) ' with my = 1.9 'mlsi gives . .' m(r_zy_1 _g)' #135 kg)'(_0 + 9.3111152) _ (a) x” '11 " ' 3400-1111311 ' ' 106m ' (35 11901911111111 + 9.8111132) - '. (b) x 1: W _ - e 12- cm _. . 3490 Nlm _ - 111551155 Why is the simmer-to (11) larger? Beceuse, just as with the . cable'in Example 4.3, the spring needs. .to provide an additional force to'accelerate the ebneretenpward. ' l' 1 ‘ 4.5 change if the helicopter were GOT IT? 4.6 (3) Would the answer to (a) in Example not at rest but moving upward at co nstant speed? (1)) Would the answer to (b) change if the helicopter were moving downward but still accelerating upward? ...
View Full Document

{[ snackBarMessage ]}