As per the given question,
Length of the uniform ladder(L)=10m
Weight of the ladder =59N
Angle between the ladder and the ground =50∘
Let the reaction on the bottom of the ladder N1 and reaction on the wall N2
Now, taking the torque about the center of the ladder
N1×2Lsin50−μN12Lcos50=N2×2Lcos50
N2=N1tan50−μN1−−−−−−−−(i)
Now, balancing horizontal force
N2=μN1−−−−−−−(ii)
From equation (i) and (ii)
2μN1=N1tan50
μ=2tan50=0.59
μ=0.59
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