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T = m (g + a) or T = M a
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limv and back down to a 15kg package on the ground. a) what is the magnitude of the least acceleration the monkey muct have if it is to ligt the package off the ground? If after the packafe has been lifted eht monkey stops its climb and holds onto the rope, what are b) the magnitude and direction of the monkey's acceleration? c) what is the tension in the rope?
I am not sure what the equations are but for c) would it be T-mg=ma? What would the other equations be?
By accelerating itself up the rope at a rate a, the monkey can apply a tension force to the rope that exceeds its weight, m g. let m be the monkey's mass amd M the package mass.
a) T = m (g + a) is the tension applied to the rope by the monkey, if "a" is measured upwards.
If the Tension is sufficient to lift the mass, T = M g
The least acceleration to move the package is given by
T = M g = m (g + a)
M/m = 1 + a/g = 3/2
Solve for a
b)After the monkey's stops climbing
T - mg = ma
T = M a
Eliminate T and solve for a.
M a = m (g +a)
M/m = 3/2 = (g/a + 1
c) Once you know a, solve for T with either equation above
1 answer
1 answer