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

Question

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. State your conversion assumption. B. The merry-go-round starts turning, accelerating very slowly. How fast is it turning when the block slips? (revs/min)

Explanation / Answer

A pyramid is at the center of a merry-go-round.

A pyramid is at the center of a merry-go-round.

The face of the pyramid is 80 degrees to vertical , i.e Angle made from horizontal is 10 degree.

sloped distance = 25.0 cm (from the center of the pyramid to the center of the block)

The coefficients of static friction = 0.300

The coefficient kinetic friction = 0.150

When the merry-go-round is not moving :

Static friction will be equal to the force required to keep the block at rest ,

i.e    Static friction = Mg*Cos(80)     where   M = 10 lbs. and g = 32.2 and Cos(80) = 0.17364818

or Static friction = 10*32.2*0.17364818 = 55.91 N .

When start moving :

For slipping

Mw2*(0.25*Cos(10))*Cos(10.) = 0.3*M*g*Cos(10)

    w2 = 0.3*9.8/(25*Cos(10))

w : unit will be rad/sec

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