# NEET Class neet Answered

**kindly provide solution to this question**

Asked by yohanagrawal | 15 Jul, 2022, 12:43: AM

Expert Answer

Let an uniform ladder AB of length l and mass m rest against the wall so that it makes angle 30

^{o}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.

N

_{1}is the normal force perpendicular to smooth wall at top contact of ladder.N

_{2}is the normal force perpendicular to floor at bottom contact of ladder.μN

_{2}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 = N

_{2 .................................(1)}At equilibrium , sum of horizontal forces equals zero

N

_{1}= μ N_{2}................................. (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 N

_{1}from eqn.(2) and rewrite above eqn.(3) asAgain let us substitute N

_{2}from eqn.(1) in above expression and after simplification we getafter 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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