We will have the following:
First, we find all the forces in x & y, that is:
[tex]\sum ^{}_{}F_x=0\Rightarrow\sum ^{}_{}F_x=F_f-N_1\colon N_1=F_f[/tex]
Here Ff is the frictional force and N1 the normal force on the point where the ladder is resting.
[tex]\sum ^{}_{}F_y=0\Rightarrow\sum ^{}_{}F_y=N_2-mg-Mg\colon N_2=(M+m)g[/tex]
Here N2 is the normal force of the resting point of the ladder on the ground. M is the mass of the person and m is the mass of the ladder.
Then we make sure that the ladder won't rotate, that is the momentum will be 0, that is:
[tex]\sum ^{}_{}\tau_i=mg\cos (\theta)(l/3)+mg\cos (\theta)(l/2)-N_2\cos (\theta)l+F_f\sin (\theta)l=0[/tex]
Here "l/3" represents the positons of the person to climb, and "l/2" represents the position of the center of mass of the ladder.
Then, we remember that we are given the static friction coefficient, and replace:
[tex]\sum ^{}_{}\tau_i=mg\cos (\theta)(l/3)+mg\cos (\theta)(l/2)-(M+m)g\cos (\theta)l+\mu_s(M+m)g\sin (\theta)l=0[/tex]
Then we divide the whole expression by cos(theta) in order to eliminate all extra trigonometric values and determie the angle:
[tex]\Rightarrow(\frac{M}{3}+\frac{m}{2})g-(M+m)g+\mu_s(M+m)g\tan (\theta)=0[/tex][tex]\Rightarrow(\frac{M}{3}+\frac{m}{2})-(M+m)+\mu_s(M+m)\tan (\theta)=0[/tex][tex]\Rightarrow\tan (\theta)=\frac{(M+m)-(\frac{M}{3}+\frac{m}{2})}{(M+m)\mu_s}\Rightarrow\tan (\theta)=\frac{(60\operatorname{kg}+10\operatorname{kg})-(\frac{60\operatorname{kg}}{3}+\frac{10\operatorname{kg}}{2})}{(60\operatorname{kg}+10\operatorname{kg})(0.50)}[/tex][tex]\Rightarrow\theta=\tan ^{-1}(\frac{45}{35})\Rightarrow\theta=\tan ^{-1}(\frac{9}{7})[/tex][tex]=\theta=52.12501635\ldots\Rightarrow\theta\approx52.1[/tex]
So, the angle would be approximately 52.1°.
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