Calculating the Net Force of Sun & Moon on Earth

[eclecticmanic]

Tides are created by the gravitational attraction of the sun and moon on Earth. Calculate the net force pulling on Earth during a)New Moon b)Full Moon c)First Quarter Moon.
Mass of the Moon: 7.35 x 10^22 kg
Mass of the Earth: 5.98 x 10^24 kg
Mass of the Sun: 1.99 x 10^30 kg
Distance from the Earth to the Moon: 3.84 x 10^8 m
Distance from the Earth to the Sun: 1.50 x 10^11 m
Universal Gravitational Constant (G): 6.67 x 10^-11
F=Gmm/d²I solved for the force of Sun on Earth
F=(6.67 x 10^-11)(1.99 x 10^30)(5.98 x 10^24)/1.5 x 10^11 = 5.29 x 10^33

And Moon on the Earth
F=(6.67 x 10^-11)(7.35 x 10^22)(5.98 x 10^24)/3.84 x 10^8 = 7.63 x 10^28

I don't know where to go from there, I think finding the solution has to do with where the moon is facing and maybe it's the difference from the distance of the sun? help me please

 

[mgb_phys]

Draw a digram with the relative positions of the earth, sun and moon.
think about how the forces form the sun and moon add

 

[eclecticmanic]

There was a picture on my worksheet. I thought that during New Moon the Sun and moon act together on Earth so I added my answers 5.29 x 10^33 + 7.63 x 10^28=5.29 x 10^33.
For Full Moon I subtracted because I though the sun and moon were working in opposite forces
5.29 x 10^33 - 7.63 x 10^28=5.29 x 10^33...I get the same answer?? misinterpreting?

For Quarter since it looked half and half on the image, I divided the sum by 2.
5.29 x 10^33/2= 2.645 x 10^33. I'm still lost.

 

[mgb_phys]

A new moon (dark) is between the Earth and sun so their gravity combines
A full moon (bright) is opposite the Earth from the sun so it pulls in the opposite direction
A half moon is 90 deg ahead of the earth

 

[Cyberman1]

I did not get that forces - did y square the distance or not?