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Asked by yohanagrawal | 15 Jul, 2022, 12:43: AM
Let an uniform ladder AB of length l and mass m rest against the wall so that it makes angle 30o
with the floor as shown in figure.

Let the electrician whose mass is 80 kg climbs on the ladder of length ( k l ) from bottom as shown in figure.
( k < 1 )

Figure shows the forces acting on the ladder . Vertically down ward forces are weight of the ladder (mg) and
weight of electrician ( Mg ) . Since the ladder is uniform , weight of ladder is acting at centre of mass of ladder.

Weight of electrician ( Mg ) is acting at a point that is at a distance ( k l ) from bottom of the ladder.

N1 is the normal force perpendicular to smooth wall at top contact of ladder.

N2 is the normal force perpendicular to floor at bottom contact of ladder.

μN2 is friction force as shown in figure that is acting at contact surface of ladder and floor, where μ is friction coefficient.

At equilibrium ,  upward vertical force equals sum of downward vertical forces

( M + m ) g = N2  .................................(1)

At equilibrium ,  sum of horizontal forces equals zero

N1 = μ N2  ................................. (2)

Let us take moment of forces by considering the pivot point as B ( bottom point of ladder )

At equilibrium , sum of counterclockwise moments equals clockwise moment .

Above expression is simplified as

............................... (3)
Let us substitute N1 from eqn.(2) and rewrite above eqn.(3) as

Again let us substitute N2 from eqn.(1) in above expression and after simplification we get

after substituting m = 40 kg , M = 80 kg and μ = ( 1/ √3 ), we get k = 1/4

Hence electrician can climb up only (1/4) of lenegth of ladder

Answered by Thiyagarajan K | 15 Jul, 2022, 08:35: AM
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