A 12 kg monkey climbs up a massless rope that runs over a frictionless tree limb

Question

A 12 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 16 kg package on the ground (see the figure). (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 (b) the magnitude and the direction of the monkey's acceleration (choosing the positive direction up), and (c) what is the tension in the rope?

Explanation / Answer

Mass of the monkey M1 = 16 kg Mass of the package M2 = 22 kg Let T = tension in the rope, a1 = upward acceleration of the monkey, a2 = acceleration with which the package moves up the ground Forces on the the monkey are 1. Tension T upward 2. Weight M1 * g downard Therefore T - M1 * g = M1 * a1--------------(1) The package moves up the ground when the normal force on it from the ground = 0 Then the forces on the package are 1. Tension T upward 2. Weight M2 * g downward Therefore T - M2 * g = M2 * a--------------(2) From equation (1), T = M1 * a1 + M1 * g From equation (2), T = M2 * a2 + M2 * g Therefore, M1 * a1 + M1 * g = M2 * a2 + M2 * g M1 * a1 = M2 * a2 + M2 * g - M1 * g a1 = (M2 * a2 + M2 * g - M1 * g)/M1-------------(3) From equation (3), for minimum value of a1, the value of a2 should be minimum. Minimum value of a2 = 0 Putting a2 = 0 in equation (3), a1 = (M2 * g - M1 * g)/M1 a1 = (M2/M1 - 1)*g = (22/16 - 1)*9.8 = (3.2 - 1)*9.8 = 2.2 * 9.8 = 21.56 m/s^2 Ans: 21.56 m/s^2 b) When the monkey holds onto the ground, then:- Let a = package's downward acceleration Then a = monkey's upward acceleration For the package, net downward force = M2 * g - T Therefore M2 * g - T = M2 * a-----------(4) For the monkey, net upward force = T - M1 * g Therefore T - M1 * g = M1 * a--------(5) From (4), T = M2 * g - M2 * a-----------------(6) From (5), T = M1 * a + M1 * g Therefore, M2 * g - M2 * a = M1 * a + M1 * g M2 * g - M1 * g = M1 * a + M2 * a (M2 - M1)*g = (M1 + M2)*a a = (M2 - M1)*g/(M1 + M2) = (22 - 16)*9.8/(22 + 16) a = 5.13 m/s^2 This is positve, which means that monkey's acceleration is in the direction in which we assumed. We assumed upward direction. Ans: 5.13 m/s^2 upward c) Putting values of a, M1 and M2 in equation (6), T = 22 * 9.8 - 16 * 5.13 = 262.3 N Ans: 262.3 N ____________________________

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