1.A light string can support a stationary hanging load of 22.6 kg before breakin

Question

1.A light string can support a stationary hanging load of 22.6 kg before breaking. An object of mass m = 3.01 kg attached to the string rotates on a frictionless, horizontal table in a circle of radius r = 0.781 m, and the other end of the string is held fixed as in the figure below. What range of speeds can the object have before the string breaks? m/s

2.A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 38.6 m. If the coefficient of static friction between crate and truck is 0.620, how fast can the truck be moving without the crate sliding?
vmax =  m/s

4.The mass of a roller-coaster car, including its passengers, is 477 kg. Its speed at the bottom of the track in the figure below is 18 m/s. The radius of this section of the track is r1 = 26 m. Find the force that a seat in the roller-coaster car exerts on a 51-kg passenger at the lowest point.

Explanation / Answer

Number 1)

The max force is mg = 22.6(9.8)

That has to be equal to the centripetal force in the horizontal circle

F = mv2/r

(22.6)(9.8) = (3.01)(v2)/(.781)

v = 7.58 m/s

Thus the speeds can go from 0 to 7.58 m/s

Number 2)

The frictional force must balance the centripetal force

Ff = Fc

Ff = uFn = umg

umg = mv2/r (mass cancels)

(.62)(9.8) = v2/38.6

v = 15.3 ms

Number 3)

Part a)

F = mv2/r + mg

F = 490(19.7)2/(10) + (490)(9.8)

F = 2.38 X 104 N

Part b)

At this point mg = mv2/r (mass cancels)

(9.8) = v2/(15)

v = 12.1 m.s

Number 4)

F = mv2/r + mg

F = 51(18)2/26 + (51)(9.8)

F = 1135 N upward

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