Exertion of force due to spin of Earth If Earth were to spin at a different speed would I be experiencing a different weight?
 A: Yes, but the effect is negligible compared to the acceleration due to gravity.
First year college physics approximations suffice to calculate the approximate effect ratio.
Firstly the acceleration due to gravity at the Earth's equator is  roughly $9.8 \frac{m}{s^2}$.
Now let's work out the approximate centripetal acceleration at the Equator.
The Earth's radius at the equator is approximately 6371 km or $ {6371e3}$ m.
Now the velocity of Earth's rotation $v$ at the equator is roughly 1670 km/hour, or , doing the unit conversions, 463.8 m/s.
Using the expression $ a_c = \frac {v^2}{\rho} $ to compute the centripetal acceleration due to the Earth's rotation (at the equator - it would be less at higher latitudes) gives a figure of 0.337 $\frac {m}{s^2}$.
So the ratio of that to the acceleration due to gravity is  0.0034, or about 0.34 percent.
Now this is a very "back of the envelope" calculation - it does not take into account effects due to the real shape of the Earth, local variations in density, and so on - but it does illustrate the approximate order of magnitude of centripetal acceleration due to the Earth's rotation relative to gravity.
The previous question Why is Earth's gravity stronger at the poles? which really is close to a duplicate has some more detailed discussion. So does Why is the Earth so fat?
A: Yes you would be experiencing a different weight, had the earth been spinning at a different speed.
Now the apparent weight of an object at a latitude $\theta$ on the earth would experience a different weight due to the change in centripetal force acting on the object.
The weight of the object is given by: $$mg’ =mg – mR \omega^2 \cos^ 2\theta$$ 
The derivation of this formula can be obtained from this link.
It can thus be concluded that, as $\omega$ of earth increases, net weight of the object decreases. This rate of decrease of weight is maximum when the object is at the equator and minimum (= 0) when object is at either pole.
That is, object weighs maximum at poles and minimum at the equator. Change in value of $\omega$ does not affect the weight if it is at poles but affects max if it is at the equator.
