Question

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 1

Explanation:

The net force on M is O as it is cancelled from all directions

if one particle is removed then the net force would be force by one m in app direction [∵ the force by m must have cancelled it when it was there]

∴F=

a

2

GHm

solution

Answer 2

Given:-

• Mass of particle at vertices= m
• Mass of particle at centre = M
• Distance between M and m=a

To find:-

1. Total force exerted on M
2. Total force on M when any one m is removed.

Solution:-

• For part a,

1. Each m will exert force on M individually
2. If we arrange the vectors of force exerted by particles on M using polygon law of addition of vectors, we find that resultant force is zero.
3. Thus force exerted on M=0

• For part b,
• One particle m is removed.
• If we arrange individual force exerted on the particle M by remaining vertices, we find that resultant is equal in magnitude to force by removed particle.
• This force can be found using gravitation formula, where we have r=a, m_1=m, m_2=M

Thus answer is $F = G\frac{m . M}{r^2}$