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 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.

 Dec 13, 2019
 #1
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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 

so 

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 

so subsituite,

Ay=1.2m

z=20N

By=1.6m

 

1.2*20=1.6*x 

24=1.6*x

x=24/1.6=15N

The force that acts on B so that the rigid rod is in equilibrium is 15N 

 Dec 13, 2019
 #2
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If the 20N force is in line with the rod, then there will be no need for any force at B, (and the reaction at the pivot will be 20N).

It's a silly question.

We need to know the angle between the rod and the line of action of the force.

 Dec 13, 2019

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