An 88 kg farmer is raising a 160 kg basket of fruit to the top of a cliff [H = 1

Question

An 88 kg farmer is raising a 160 kg basket of fruit to the top of a cliff [H = 120 m]. He rides a 640 kg yak and uses a rope with a length of 140 m that passes over a small frictionless pulley. The yak's hooves and the Himalayan rock share a kinetic friction coefficient of 0.22. The rope is attached to the yak's back at a point that is 1.8 m from the ground. The drawing is not to scale. When the fruit reaches the top of the cliff, the farmer dismounts from the yak to unload the basket. To his surprise, the yak begins to slide towards the cliff edge as the basket descends. Calculate the acceleration of the yak. How fast is the basket moving at the instant it reaches the bottom of the cliff? Does the yak slide off the edge of the cliff? Support your answer with calculations.

Explanation / Answer

a) Refer above figure,

Applying Newton’s second law to m1, T-Fk=m1a

T-km1g=m1a

T=km1g+m1a ---------------(1)

Applying Newton’s second law to m2,

m2g-T=m2a

from(1)

m2g- km1g-m1a=m2a

m1a+m2a= m2g- km1g

a= (m2g- km1g)/(m1+m2) = (160*9.8-0.22*640*9.8)/(160+640) = 0.2352 m/s^2

since for m2 acceleration is negative a= -0.2352 m/s^2

b) Use kinematic equation,

Vf^2=Vi^2+2*a*d =0^2+2*-0.2352*120        =>   Vf= -7.5 m/s

c) No yak will not slide off the cliff since height of the cliff is 120m , so 120m rope will be down the cliff , but another 20m rope is remained above the cliff free as soon basket reaches at the bottom since total rope is 140m.

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