Answer:
23. 63 degrees
Explanation:
Let us draw a free body diagram.
The net forces on ladder + painter must equal zero. Therefore, along the y-direction:
[tex]N_{}-m_lg-m_p_{}g=0[/tex][tex]\Rightarrow N=(m_l+m_p)g[/tex]
Along the x-direction
[tex]F_r-F_1=0[/tex]
Now, we calculate the torque about the about O.
The torque is must be zero:
[tex]-m_pgd\cos _{}\theta-m_lg(\frac{l}{2})\cos \theta+N\cos \theta+F_r\sin \theta=0[/tex]
simplifying the above gives
[tex]F_r\sin \theta=(m_pd+m_l(\frac{l}{2})-\frac{N}{g})g\cos \theta[/tex]
Since
[tex]F_r=\mu N=\mu(m_l+m_p)g[/tex]
the above becomes
[tex]\mu(m_l+m_p)g\sin \theta=(m_pd+m_l(\frac{l}{2})-\frac{N}{g})g\cos \theta[/tex][tex]\Rightarrow\mu(m_l+m_p)\sin \theta=(m_pd+m_l(\frac{l}{2})-\frac{N}{g})\cos \theta[/tex]
dividing both sides by cosine gives
[tex]\mu(m_l+m_p)\frac{\sin\theta}{\cos\theta}=(m_pd+m_l(\frac{l}{2})-\frac{N}{g})[/tex]
further division gives
[tex]\frac{\sin\theta}{\cos\theta}=(m_pd+m_l(\frac{l}{2})-\frac{N}{g})\cdot\frac{1}{\mu(m_l+m_p)}[/tex]
substitute the value of N and we get:
[tex]\frac{\sin\theta}{\cos\theta}=(m_pd+m_l(\frac{l}{2})-\frac{(m_l+m_p)g}{g})\cdot\frac{1}{\mu(m_l+m_p)}[/tex][tex]\frac{\sin\theta}{\cos\theta}=(m_pd+m_l(\frac{l}{2})-(m_l+m_p))\cdot\frac{1}{\mu(m_l+m_p)}[/tex]
putting in m_p = 60 kg, m_l = 10 kg, l = 6.0 m and d = 1/3 * 6 = 2, and u = 0.5 gives,
[tex]\frac{\sin\theta}{\cos\theta}=(60\cdot2+10(\frac{6}{2})-(10+60))\cdot\frac{1}{0.5(10+60)}[/tex][tex]\Rightarrow\frac{\sin \theta}{\cos \theta}=\frac{16}{7}[/tex][tex]\Rightarrow\tan \theta=\frac{16}{7}[/tex]
taking the inverse tan of both sides gives
[tex]\theta=\tan ^{-1}(\frac{16}{7})[/tex][tex]\theta=66.37^o[/tex]
Therefore, the required angle against the wall is
[tex]\alpha=90-\theta=23.63^o[/tex]
Hence, the ladder should lean 23.63 degrees against the wall.
