Consider the situation as shown in the following diagrams. Find net gravitational force on the particle of mass `m`.

Explanation

(a) Gravitational force on `m`,
Due to `M, F_(1)=(GMm)/((2d)^(2))`, along `CA`
Due to `2M, F_(2)=(G.2Mm)/((d)^(2))`, along `CA`
Resultant force on `m`
`F=F_(1)+F_(2)=(GMm)/(d^(2))(1/4+2)`
`=9/4(GMm)/(d^(2))`, along `CA`
(b) Force on `m`
Due to `36M, F_(1)=(G.36 Mm)/((6d)^(2))=(GMm)/(d^(2))` along `BA`
Due to `64 M, F_(2)=(G.64 Mm)/((8d)^(2))=(GMm)/(d^(2))` along `CB`

`F=sqrt(F_(1)^(2)+F_(2)^(1))=(sqrt(2)GMm)/(d^(2))`
`theta=45^(@)`
(c)
The force on `m` due to `M` and due to `M` at `C` are equal and opposite.
The foece on `m` due to `M` at `A`
`F=(GMm)/((sqrt(3)d//2)^(2))=4/3 (GMm)/(d^(2))`, along `DA`
(d) `F_(0)=(GMm)/(d^(2))`
Resultant force on `m`


`F=2F_(0)cos 30^(@)=sqrt(3) F_(0)=(sqrt(3)GMm)/(d^(2))`
(c)
`F_(0)=(GMm)/(d^(2))`
`F_(1)=(GMm)/((sqrt(3)d//2)^(2))=4/3 (GMm)/(d^(2))`

Resultant force on `m`
`F=2F_(0)cos30^(@)+F_(1)`
`=sqrt(3)F_(0)+F_(1)`
`=(sqrt(3)GMm)/(d^(2))+4/3 (GMm)/(d^(2))`
`=(sqrt(3)+4/3)(GMm)/(d^(2))`, along `CD`
(f)
Force on `m`,
Due to `2m,F_(1)=(G.2m.m)/(d^(2))`, along `AB`
Due to `3m,F_(2)=(G.3m.m)/((sqrt(2)d)^(2))`, along `AC`
Due to `4m,F_(3)=(G.4m.m)/(d^(2))`, along `AD`

`F_(x)=F_(3)+F_(2)cos45^(@)=(Gm^(2))/(d^(2))(4+(3)/(2).(1)/(sqrt(2)))`, along `AD`
`F_(y)=F_(1)+F_(2)cos45^(@)=(Gm^(2))/(d^(2))(2+(3)/(2).(1)/(sqrt(2)))`, along `AB`

`F=sqrt(F_(x)^(2)+F_(y)^(2))`
`tan theta=(F_(y))/(F_(x))`