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Safari File Edit View History Bookmarks Window Help edugen.wileyplus.com Google sa kb USA. Ch Wiley PLUS The mine skip is being hauled to the surface over the curved track by the cable wound around the 33- in. drum, which turns at the constant clockwise speed of 144 rev/min. The shape of the track is designed so that y 3x2/50, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 3.7 ft below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by dy 2 13/2 dx2 33" Answer: a ft/sec the tolerance is +/-2% Click if you would like to Show Work for this question Open Show Work 27 Mechanical Engineering question I Chegg.com

Explanation / Answer

given , y = x^2/25

dy/dx = 2*x/50

= x/25

d^2y/dx^2 = 1/25

so, radius of curvature, rho = (1 + (1/25)^2)^1.5/(1/25)

= 25 feet.

diameter of the drum, a = 33 inches

= 33*(1/12) feet

= 2.75 feet

angular speed, w = 144 rev/min

= 144*2*pi/60

= 15.07 rad/s

so, v = (d/2)*w

= (2.75/2)*15.07

= 20.72 feet/s

so, a = v^2/rho

= 20.72^2/25

= 17.2 feet/s^2

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