A uniform ladder of length 10.0 m and mas 16.0 kg is resting against a vertical wall making an angle of `37^@` with it. An electrician weighing 60.0 kg climbs up the ladder. If the stays on the ladder at a point 8.00 m from the lower end, will be normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work safely?

Explanation

If the electrician works safetly, the ladder `+` man system should be in equilibrium

For equilibrium `SigmaF=0` and `Sigmatau=0`
for `SigmaF_(x)=0, N_(a)=f`
for `SigmaF_(Y)=0, N_(B)=mg+Mg=60xx10+16xx1=760N`
Taking torque about `B`,
`N_(A)10 cos 37^@=mg8sin37^@+Mg5sin37^@`
`N_(A)=(tan37^@)/10(65xx10xx8+16xx10xx5)`
From i `f=N_A=420N`
For maximum value of coefficient of friction.
`flemuN_(B)`
`mugef/N_(B)=420/760=0.55`