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Grade: 12th pass

                        

a ladder of length l and mass m is placed against a smooth vertical wall but ground is not smooth coefficient of friction is ‘t’,then for what angle will it stay in equilibrium?

5 years ago

Answers : (1)

Mihir
69 Points
							
In this problem, the body must be in torsional/ rotational and translational equilibrium. Now let’s assume two normal forces to be operative, namely N and N’. Let N act perpendicular to the wall and N’ act perpendicular to the floor. Now the frictional force will act along the direction towards the wall, the magnitude being kN’ where k is the coefficient of friction.
Now to establish a translational equilibrium, the weight must be balance by N’ and the frictional force must be balanced by N.
Therefore we have the following mathematical relations:
mg= N’…... 1
kN’= N........2
To establish a rotational equilibrium, we need to understand that the torque produced by N’ is clockwise and the torque produced by N and frictional force is anticlockwise. Further on conserving it about centre of mass, we get the following equations:
N’ x L/2 x cos∂= Nx L/2 x sin∂ + kN’x L/2 sin∂
now using the above two realtions,
tan∂= 1/2k
 
5 years ago
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