show the acceleration due to gravity affected at a latitude due to the rotational motion of earth
Answers
If we ignore extremely minor effects due to general relativity and your frame of reference relative to a rotating body, and special relativity effects due to very fast spinning bodies (neither of which are significant on Earth), then the answer is none at all - directly. But indirectly there are some effects.
Before we get to the real Earth, first lets approximate things by considering Earth as a rigid sphere. The acceleration due to gravity is g=GM/R^2. Now we start to spin the earth, but its shape and density distribution does not change because it is rigid. The acceleration due to gravity is g=GM/R^2 - unchanged!
But in addition to gravity there is also another force at work due to the rotation of the sphere, and you have to add that acceleration on to the acceleration due to gravity to get the net acceleration on your body sitting on the surface.
This rotational force is very real, and its cause is not at all intuitively obvious. But any time you get a change in a velocity vector, you can know for certain from Newton's laws that a force has caused it, so it is real.
Because the velocity vector is changing on the equator of a rotating sphere (its direction is changing but not its magnitude) we can infer two things: one, a force is acting, and (because only the direction is changing) that force is acting perpendicular to the velocity vector. And experiment confirms the magnitude of the force at the equator is m.v^2/R, and so the acceleration on your body is a=v^2/R (outwards from the spheres rotational axis) As you move to one of the rotational poles, closer to the rotational axis, v goes to zero, and so acceleration due to this rotation also goes to zero as per the formula. (It helps to understand that tangential velocity v is related to rotation rate, omega, by v=omega * r where r is the distance from the rotational axis)
As you can see from the formula this acceleration due to a force of rotation has nothing at all to do with gravity - it is in addition to gravity, and the acceleration due to gravity remains unchanged, making the net acceleration on your body at the equator of a=GM/R^2-v^2/R.
So yes the weight force we experience at the equator is a bit less than the weight force due to gravity, but not at all because gravity has changed value; there is just another force we have to account for as well as gravity.
Now to the real Earth. It is no accident the acceleration (inwards to the centre due to gravity) is greater than the acceleration outwards due to rotation, leaving a positive (directed towards the centre) result so we do not go flying off into space and can sit comfortably in our armchair, relying on that good old electromagnetic force to make up the difference to a=0 so we do not plummet through the ground on our "Journey to the Centre of the Earth".
If the opposite was true, we would have to hang on for dear life, and hope that what we were hanging on to, was also hanging on to something else, all the way down to the centre. In practice if this were so, the whole Earth would fly apart, because its material properties are much weaker in tension than compression.
And speaking of indirect effects, these material properties do come in to play on the real Earth. As the rotational speed increases, things move around,because the earth is not rigid - it will bulge more at the equator, and will stratify by density a bit more, and so on. So indirectly R will change, as will density distribution, both of which means the value of g will in fact change; but only as an indirect effect of rotation, not a direct one.