a pyramid is at the center of the merry go round the face of

a pyramid is at the center of the merry go round. the face of the pyramid is 80 degrees to vertical. a block that weighs 10 lbs sits on one of the sides of the pyramid, at a sloped distance of 25.0 cm. (from the center of the pyramid to thecenter of the block) the coefficent of its static and kinetic friction are .300 and .150 respectively. A) the merry go round is not moving, what is the static friction force in Newtons?

B) the merry go round starts turning, accelerating very slowly. How fast is it turning when the block slips? __ revs/min

Solution

A) perpendicular to the face:

N - mgsin80 = 0

m = 10 x 0.454 kg = 4.54 kg

N = 4.54 x 9.8 x cos10 = 43.82 N

Maximum static fricrtion = usN = 0.3 x 43.82 = 13.14 N

along the surface,

f - mgcos80 = 0

f = 4.54 x 9.8 x cos80 = 7.73 N which is withing permitted value of static friction.

hence value of static friction = 7.73 N

B) when total force along the surface will reach the maximum value of friction then block
will start to move.

perpendicular to the surface,

N - mgsin80 + mw^2 r sin10 = 0

N = mgsin80 - mw^2rsin10

fmax = 0.3 ( mgsin80 - mw^2rsin10 )

along the surafce,

fmax - mgcos80 - mw^2rcos10 = 0

0.3 ( mgsin80 - mw^2rsin10 ) - mgcos80 - mw^2rcos10 = 0

0.3 w^2 r sin10 + w^2 r cos10 = 0.3 x g x sin80 - g cos80

0.259w^2 = 1.19

w = 2.15 rad/s

in rev/min, w = 2.15 (1/2pi rev) /(1/60 min) = 20.50 rev/min