2. A pyramid is at the center of a merry-go-round. The face of the pyramid is 80

Question

2. A pyramid is at the center of a 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 the center of the block). The coefficients of static and kinetic friction are 0.300 and 0.150, 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? (In revs/minute)

Explanation / Answer

part A : Fn = mg cos theta and Ff = mg sin theta

so

us Fn = mg sin theta

FF = 10 * 32 * 0.17

Ff = 55.57 is the frictional force

------------------------------------

here Force N = mg cos theta + Fc sin theta

and

Fc cos theta = Ff - mg sin theta

so

Fc = u mg - mg tan theta /(1 - uk tan theta)

Fc = (0.15 * 10 * 32) -(32* 10 * tan 10)/(1 - (0.15 * tan 10)

Fc = 8.38 N

this is equal to centripetlal froce = mv^2/R

v^2 = 8.38 r/m

v = 8.38 * 0.25 * 3.28/10

v = 0.83 f t/s

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