Five particles each of mass 'm' are kept at five verties of a regular pentagon . A sixth particles of mass 'M' is kept at center of the pentagon 'O'.Distance between'M' and is 'm' is 'a'. Find
(a) net force on 'M'
(b) magnitude of net force on 'M' if any one particle is removed from one of the verties.

Answer

A, B

Explanation

(a)`F_A = F_B = F_C = F_D = F_E = (GMm)/(a^(2))=F`(say)
Angle between two successive force vectors is `theta = (360^(@))/(5)=72^(@)`.
When these five force vectors are added as per polygon law of vector addition we get another closed regular polygon as shown below.


Therefore, net resultant force on 'M' is zero. (b) When particle at A is removed, then `F_(A)` will be removed. So, magnitude of net force will be `F` as shown below:
`:.F_("net") = F =(GMm)/(a^(2))`