Free body diagram for person weighing book at the north pole and at equator

Applying Newton’s 2nd law of motion at the north pole;

ΣF = 0

F_NP - F_g = 0

F_NP = F_g

F_NP = Weight of book at pole = W_P = mg

W_P = (10 kg)(9.8 m/s²) = 98 N

Thus the apparent weight of book at north pole (W_P) is 98 N.

Now finding apparent weight of the book on equator:

The net force acting on it is centripetal force i.e., ΣF = F_c

F_c = -ma_c

where ac is centripetal acceleration

and a_c = v² / r

Thus, F_c = -(mv²) / r

From free body diagram, Fc = F_NE - F_g

F_NE - F_g = - (mv²) / r

F_NE =F_g - (mv²) / r

F_NE = mg - (mv²) / r

Tangential velocity is defined by, v = rw. Thus

F_NE = mg - mrw² ………. (1)

Here w = ?

The angular velocity is that of the equator so it can be found as;

w = 2π / T

w = 2π rad / (24 x 3600) s

w = 7.27 x 10^-5 rad/s

Also the radius of earth = r = 6.4 x 10⁶ m

Substitute values of m, g, r, and w in eq(1)

F_NE = mg - mrw²

F_NE = (10 kg)(9.8 m/s²) - (10kg)(6.4 x 10⁶ m)(7.27 x 10^-5 rad/s)²)

F_NE = Weight of book at equator = W_E = 97.66N

Thus the weight of the book at the equator (W_E) is

ΔW = W_P - W_E = 98 N - 97.66 N = 0.34 N

Thus scale reading at the north pole is greater than at the equator by 0.34 N

A 5.0 kg object is attached to one end of a rope, which has negligible mass. The object moves in a uniform circular motion in a vertical circle of radius 2.0 m. Calculate the period of revolution, if the tension in the rope is zero at the topmost point.