A rigid rod AB length 2.8 m of negligible mass is acted upon by a force of 20 N at end A. If the rod is pivoted at 1.2 m from end A, find the value of the force at B such that the rod is in equilibrium and the reaction at the pivot.
For an object to be in equilibrium, resultant of all moments(Principal of moments) must be equal (I.e anti clockwise moments= clockwise moments) & Resultant of all forces (Fres)=0 as well.
AB is the whole length of the rigid rod which is 2.8
Imagine point A at one end and a force acting on it downwards 20N (There is also a contact/reactant/normal force acting on it on the opposing direction so that sets the Fres=0) "If the rod is pivoted at 1.2m from end A" Well the whole length of the rigid rod is 2.8, away from A 1.2m where is the pivot so the pivot away from point B would be 2.8-1.2=1.6m
Now let the pivot be y
let force that acts on A be z
let force that acts on B be x
Ay*z must be equal to By*x
we know the distance Ay and By also we know z but we want to find x so that Ay*z=By*x
The force that acts on B so that the rigid rod is in equilibrium is 15N