Question

a ladder 28 m long of uniform mass distribution in inclined at 45 degree on a vertical wall the coffecient of friction of ladder and wall and ladder and floor is 1/3 and 1/2

Answer 1

Explanation:

length of ladder =5 m

distance of base form vertical wall =3 m

mass of ladder =30 kg=m

friction act only on point A

let friction force at

A=f

r

from figure

cosθ=

5

3

θ=53

o

making free body diagram of ladder

Then for vertical balance

=N

1

=mg.....(1)

for horizontal balance

fr=N.....(2)

Taking rotational equilibrium about C

as torque is only due to perpendicular component

⇒AC=BC=

2

5

AC×fr sinθ+BC×N sinθ=AC×N

1

cosθ

=

2

5

×frsin53+

2

5

×frsinθ=

2

5

×mg cosθ

as (N=f

r

,N

1

=mg from (1) & (2))

=2f

r

sin53=30×10cos53

=fr=

2

30×10

×

3

4

=200 N

Hence friction between ladder and ground is 200 N