A uniform ladder of length 10.0 m and mass 16.0 kg rests against a...

Let $N_g$ be the normal force exerted by the ground, $f$ be the frictional force, and $N_w$ be the normal force exerted by the frictionless wall.

Balancing the vertical forces:

$N_g = m_l g + m_e g$

$N_g = (16.0 + 60.0) \times 9.8 = 76.0 \times 9.8 = 744.8 \text{ N} \approx 745 \text{ N}$

Balancing the horizontal forces:

$f = N_w$

Taking torque about the lower end of the ladder. The ladder makes an angle of 37^{}^{{}^{\circ}} with the vertical wall, so the perpendicular distances for the weights are proportional to \sin(37^{}^{{}^{\circ}}) and for the normal force from the wall is proportional to \cos(37^{}^{{}^{\circ}}):

N_w L \cos(37^{}^{{}^{\circ}}) = m_l g \left(\dfrac{L}{2}\right) \sin(37^{}^{{}^{\circ}}) + m_e g d \sin(37^{}^{{}^{\circ}})

N_w \times 10.0 \times \cos(37^{}^{{}^{\circ}}) = 16.0 \times 9.8 \times 5.0 \times \sin(37^{}^{{}^{\circ}}) + 60.0 \times 9.8 \times 8.0 \times \sin(37^{}^{{}^{\circ}})

N_w \times 10.0 = (784 + 4704) \tan(37^{}^{{}^{\circ}})

$N_w \times 10.0 = 5488 \times \dfrac{3}{4} = 4116$

$N_w = 411.6 \text{ N}$

Thus, the force of friction is $f = 411.6 \text{ N} \approx 412 \text{ N}$.

The minimum coefficient of friction $\mu$ required for the electrician to work safely is:

$\mu = \dfrac{f}{N_g} = \dfrac{411.6}{744.8} \approx 0.553$