g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride

2.52 rpm

Explanation:

given that

diameter of the wheel, d = 25 m

Mass of the person, m = 75 kg

Weight experienced, N = 810 N

Since diameter is 25, radius then is 25/2 = 12.5 m

We all know that,

v = rw

Also, the passengers weight is equal to the centripetal acceleration, and thus

mg = mv²/r

Substitute for v, we have

mg = m/r * (rw)²

mg = mr²w²/r

g = rw²

If we make w the subject of formula, we have

w² = g/r

w = √(g/r)

mg = 810

75 * g = 810

g = 810 / 75

g = 1.08 m/s²

w = √(g/r)

w = √(1.08 / 12.5)

w = √0.0864

w = 0.294 rad/s

Since the question asked us in rpm, we convert to rpm

0.294 * (60 / 2π)

2.52 revolution per minute.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards