Answer:
The max distance is [tex]z= 2.588 m[/tex]
Explanation:
A sketch of this question is shown on the first uploaded image
From the question we are told that
The length of the ladder is [tex]L = 3m[/tex]
The angle with respect to the horizontal is [tex]\theta = 50^o[/tex]
The coefficient of static friction is [tex]\mu = 0.55[/tex]
The mass of sally is [tex]m_s = 60kg[/tex]
The mass of ladder is [tex]m_l = 80kg[/tex]
The frictional force acting between the ladder and the ground is mathematically represented as
[tex]F_F = \mu (m_sg + m_l g)[/tex]
Where g is the acceleration due to gravity
Substituting values
[tex]F_F = 0.55(60*9.8 + 80*9.8 )[/tex]
[tex]= 754.6N[/tex]
For the ladder not to slipping the frictional force must be equal to the Normal force
Which implies that the normal force [tex]F_N[/tex] [tex]= 754.6N[/tex]
Let assume that sally is at a distance z from point B as shown in the diagram
Form the ladder to slip then the net torque about must be equal to zero
The net torque is mathematically represented as
[tex]F_N L sin \theta - m_lg*\frac{L}{2} cos \theta - m_s g z cos \theta = 0[/tex]
Making z the subject
[tex]z = \frac{F_N L sin \theta - m_l * g * \frac{L}{2} cos \theta }{m_s g cos \theta}[/tex]
[tex]= \frac{754.6 *3 sin 50 - 80 * 9.8 *1.5\ cos 50 }{60 * \ 9.8 \ cos 50}[/tex]
[tex]z= 2.588 m[/tex]
