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.

a) 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.

b) How fast is the basket moving at the instant it reaches the bottom of the cliff?

c) Does the yak slide off the edge of the cliff? Support your answer with calculations.

Explanation / Answer

angle of incline above horizontal,

@ = tan^ -1( 120/140) = 40.60 degrees ,

On yAk in vertical,

N = Mg

friction = uN = 0.22Mg

along Horizontal

T - 0.22Mg = Ma

on basket ,

mg - T = ma

adding them

mg - 0.22Mg = (m + M)a

160g - 0.22x640g = (160 + 640)a

a = 0.235 m/s^2 .......Ans

b) basket will tarvel 120 m with 0.235 m/s^2

Using v^2 - u^2 = 2aH

v^2 - 0 = 2 x 120 x 0.235

v = 7.52 m/s

c) after basket reaches to botttom.

distance left between yak and cliff end d =140 - 120 - 1.8 =18.2 m

now deccelaration will be a = - ug= -0.22 x9.81 = -2.16 m/s2

using v^2 - u^2 = 2ad

0 - 7.52^2 = 2 x -2.16 x d

d = 13.10 m

so yak will slide 13.10 m . (less than 18.2m )

so it will be safe.

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