Question

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a point at the poles experiences no centripetal acceleration. (a) Show that at the equator the gravitational force on an object must exceed the normal force required to support the object. That is, show that the object's true weight exceeds its apparent weight. (b) What is the apparent weight at the equator and at the poles of a person having a mass of 75.0 kg

Answer 1

Answers:

a) $F_{g}=735 N$ and $n=732.47 N$, hence $F_{g} > n$

b) $n_{poles}=735 N$  $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the centripetal force $F_{c}$ and the gravitational force $F_{g}$ (also called true weight) are directed "downward", while the normal force $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation:

$n_{equator}-F_{g}=-F_{c}$ (1)

Where:

$F_{g}=m g$ being $m=75 kg$ the mass and  $g=9.8 m/s^{2}$ the acceleration due gravity

$F_{c}=m a_{c}$ being $a_{c}=0.0337 m/s^{2}$ the centripetal acceleration at the equator

According to this (1) is rewritten as:

$n_{equator}-mg=-m a_{c}$ (2)

Isolating $n_{equator}$:

$n_{equator}=-m a_{c} + mg$ (3)

$n_{equator}=m(-a_{c}+g)$ (4)

$n_{equator}=75 kg (-0.0337 m/s^{2}+9.8 m/s^{2})$ (5)

$n_{equator}=732.47 N$ (6) This is the apparent weight at the equator

The true weight is given by $F_{g}=m g=75 kg (9.8 m/s^{2})$

Hence:  $F_{g}=735 N$ (7)

As we can see  $F_{g} > n_{equator}$

b) Now we have to calculate the apparent weight at the poles $n_{poles}$:

$n_{poles}-F_{g}=-F_{c-poles}$ (8)

Since $F_{c-poles}=0$ (8) is rewritten as:

$n_{poles}=F_{g}$ (9)

$n_{poles}=m g$ (10)

$n_{poles}=(75 kg)(9.8 m/s^{2})=735 N$ (11)

So, the apparent weight of the person at the poles is 735 N and at the equator is 732.47 N