thenFN= 0 and ap= 0. According to the second law equation for the package, this means F= mpg. Substituting mpgfor Fin the equation for the monkey, we solve for am:()2215 kg 10 kg 9.8 m/s4.9 m/s .10 kgpmmmmmgFmga−−−====(b) As discussed, Newton’s second law leads to pppgma−=for the package and mFmgmafor the monkey. If the acceleration of the package is downward, then the acceleration of the monkey is upward, so am= –ap. Solving the first equation for Fgam=+=−dibgand substituting this result into the second equation, we solve for am:2215 kg10 kg 9.8 m/s2.0 m/s .15 kg10 kgmga−−=++(c) The result is positive, indicating that the acceleration of the monkey is upward. (d) Solving the second law equation for the package, we obtain ()2215 kg 9.8 m/s2.0 m/s120N.Fmga=−=57. We take +yto be up for both the monkey and the package. (a) The force the monkey pulls downward on the rope has magnitude
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.