Consider a body having centre of mass at (x,y) and many forces acting on it then why it is said that to consider the the force acting on centre of mass then solve the problem why is that true
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In some scenarios it is a useful shortcut to replace all of the forces acting on a body by a single net force acting on its centre of mass. This can be useful if there is a force that is distributed across the whole body, such as an object's weight. We know that weight actually acts on every individual part of an object, but for the purposes of determining the rate of change of momentum of the object, we can replace the weights of each part of the object by a single net weight $mg$ acting on its centre of mass.
However, this shortcut does not work in all scenarios. In particular, it is only useful for rigid bodies i.e. if the relative positions of all parts of the body are fixed. Another restriction is that it tell us nothing about the rotational motion of the body - if a couple (a pair of equal and opposite forces) acts on the body then we know that the linear momentum of the body is constant because the net force acting on it is zero, but the couple can still change the angular momentum of the body.
gandalf61
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The answer lies in the definitions of the quantities involved in your question. Throughout my answer, by 'particle' I refer to 'point particle'. The centre of mass of a body is the mass weighted average of the constituent particles of your (supposedly) rigid body. $\vec{R_{\rm CM}} = \frac{\Sigma_{i}m_i\vec{r_i}}{\Sigma_{i} m_i}$. Take time derivative of this quantity and you have: $\dot{\vec{{R}_{\rm CM}}} = \frac{\Sigma_{i}m_i\dot{\vec{{r}_i}}}{\Sigma_{i} m_i}$ It is important to note here that the denominator is just the total mass of the body and the numerator in the above equation consists of sum of momenta of all the particles that constitute the body. Thus, momentum being a vector, and by the definition of a rigid body (that the relative positions of all the particles remain fixed wrt each other and that information of any change in position travels instantaneously between them), we have that the momentum of the entire body is the sum of momenta of individual particles that consist the body. So, we basically have upon rearranging $\vec{P} = \Sigma_{i} \vec{p_i}$, where LHS denotes the momentum of the rigid body. Take time derivative on both sides and it is seen that $\vec{F} = \Sigma_{i}\vec{F_i}$. Hence, considering the forces acting on the center of mass solves all problems regarding the translational motion of a rigid body, as it is, by definition equivalent to solving for the motion of each particle that constitute it and then taking into account the overall motion.
