Question

A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it makes an angle of 1.16 radians with the floor. A man weighing 734 N climbs slowly up the ladder. When he has climbed to a point that is 7.46 m from the base of the ladder (measured along the ladder), the ladder starts to slip. Find the coefficient of static friction between the floor and the ladder.

Answer 1

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

$\tau_{net} = 0$; and

$-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0$

However, by rearrangement;

$fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta \\ \\ \mu(m_L + m) gL \ sin \theta = (323 \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0$

$\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}$

$\mathbf{\mu= 0.3625 }$