Question

4 particles eaxh of mass m r placed at vertices of a square of side a..Find net force acting on any one particle...​

Answer 1

According to pythagors theorem
In △ACD
(AC)
2
=(AD)
2
+(DC)
2

⇒ (AC)
2
=a
2
+a
2

⇒ Ac=a
2
​

Now,
we know that
F=
r
2

Gm
1
​
m
2
​

​
[gravitational force]
⇒ F
AB
​
=
a
2

Gm×m
​
=
a
2

Gm
2

​

F
AD
​
=
a
2

Gm
2

​

F
AC
​
=
(AC)
2

Gm
2

​
=
2a
2

Gm
2

​

F
net
​
=
(F
AB
2
​
+(F
AD
2
​
))
​

=
(
a
2

Gm
2

​
)
2
+(
a
2

Gm
2

​
)
2

​

⇒ F
net
​
=
2
​

a
2

Gm
2

​

tanθ=
F
AD
​

F
AB
​

​

⇒ tanθ=1
⇒ θ=45
o

hence
F
net
​
and F
CD
​
is in same
direction.
⇒ Net force on particle A=F
net
​
+F
CD
​

2
​

a
2

Gm
2

​
+
2a
2

Gm
2

​

=
a
2

Gm
2

​
[
2
​
+
2
1
​
]
=
2a
2

Gm
2

​
(
2
​
+


Hope it’s help you

Answer 2

$f(a) \: = \: g \frac{m1 \times m2}{ {r}^{2} }$

$= 24 \frac{ {m}^{2} }{a}$

Similarly, f(c) is also the same. So it f(a) and f(b).

Hence,

$f(a) = f(b ) = f(c) = 24 \frac{ {m}^{2} }{a}$

Hence the net force acting on each particle is 24 { m^2 / a }