A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground (Fig. 5-54). (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If,after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey’s acceleration and (d) the tension in the rope?

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Solution

We take +y to be up for both the monkey and the package. The force the monkey pulls downward on the rope has magnitude F. According to Newton’s third law, the rope pulls upward on the monkey with a force of the same magnitude, so Newton’s second law for forces acting on the monkey leads to
$ F – m_{mg} = m_ma_m $,$ where mm is the mass of the monkey and am is its acceleration. Since the rope is massless $F = T$ is the tension in the rope. The rope pulls upward on the package with a force of magnitude F, so Newton’s second law for the package is
$F + F_N - m_pg = m_pa_p $,$ where mp is the mass of the package, $a_{p}$ is its acceleration, and $F_{N}$ is the normal force exerted by the ground on it. The free-body diagrams for the monkey and the package are shown to the right (not to scale). Now, if F is the minimum force required to lift the package, then $F_{N} = 0$ and $a_{p} = 0$ . According to the second law equation for the package, this means $F = m_{p} g$ .
(a) Substituting mpg for F in the equation for the monkey, we solve for $a_{m}$ :
$a_{m} = \frac{F - m_{g} g}{m_{m}} = \frac{(m_{-} m_{m}) g}{m_{m}} = \frac{(15 k g - 10 k g) (9.8 m / s^{2})}{10 k g} = 4.9 m / s^{2}$
(b) As discussed, Newton’s second law leads $F - m_{p} g = m_{p} a_{p}^{'}$ for the package and $F - m_{p} g = m_{m} a_{p}^{'}$ for the monkey. If the acceleration of the package is downward, then the acceleration of the monkey is upward, so $a_{m}^{'} = - a_{p}^{'}$ . Solving the first equation for F
$F = m_{p} (g + a_{p}^{'}) = m_{p} (g - a_{m}^{'})$
and substituting this result into the second equation
$m_{p} (g - a_{m}^{'}) = m_{p} (g - a_{m}^{'})$
we solve for $a_{m}^{'}$ :
$a_{m}^{'} = \frac{(m_{p} + m_{m}) g}{m_{p} + m_{m}} = \frac{15 k g - 10 k g}{(9.8 m / s^{2})} 15 k g + 10 k g = 2.0 m / s^{2}$
(c) The result is positive, indicating that the acceleration of the monkey is upward.
(d) Solving the second law equation for the package, the tension in the rope is
$F = m_{p} (g - a_{m}^{'}) = (15 k g) (9.8 m / s^{2} - 2.0 m / s^{2}) = 120 N$ .

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