Question

MY NOTES A mobile is constructed of light rods, light strings, and beach souvenirs as shown in the figure below. Ifm - 12.09, find values (ing) for the following. (Let d, -4.50 cm, d, - 5.00 cm, dy 10 cm, de-5.60 cm, ds - 2.60 cm, and do -3.50 cm.) 4 des 9 9 9 (d) What I? If m, accidentally falls off and shatters when it strikes the floor, the rod holding m, will move to a vertical orientation so that my hangs directly below the end of the red supporting my To what values should m, and my be adjusted so that the other two rods will remain in equilibrium and be oriented horizontally? (Enter your answers in g.) 9

Answer 1

Given information mass m4 12 goum dimensions d, = 4.80 cm d2 = 5.90 cm d3 = 1.80 cm d4 = 5.60 cm 2.60 cm do = 3.50 cm. ds - (Q2l For the lowest zod; under the equilibrium condition, m4.dg = m,.de mi m4. d5 do .. mu (12) (2.60) (3.50)

:. mi = 8.91 8.91 grum 62 For the middle rod : under the equilibrium condition, m, .dz = (m.tmw.d4 .. me 1 cm, + m4). 04 dz (8.91+12) (5.60) (1.80) m₂ = . m₂ 65.053 gram coFor the top. most rod : under the equilibdium condition mz.da (mit matme).d. m₂ (m, + ma + m₂). di i m₃ (8.91+ 12 + 65.053).(4.80) (5.90)

:: m2 = 69.94 goum (d) Now mass m, accidentully falls off, and the zod (lowest) holding my move to a vertical osientation for the middle rod : m₂ .dz m4.d4 m2 = m4. d4 d₃ .. m2 = (12).(5.60) (1.80) m2 = 37.33 grum For the top-most ood: m3.02. (m, + m4.).d. :: m3 (M2+ma). di dz

m₂ = (37.33+ 12). (4.80) (5.90) i m3 = 40.135 gaum Answer: (0) me = 8.91 g Ch.) m2 6 5.053 op m3 69.949 co) mg = 37.339 mz 40.1359