An 8.0 m, 200 N uniform ladder rests against a smooth wall.The coefficient of static friction between the ladder and theground is 0.60, and the ladder

This is a bit challenging. There are three equations you must write. Three things will also be unknown to you. Sum of horizontal forces, Sum of vertical forces, and Sum of torques are the three equations. The solution involves placing the person "x" steps up the ladder, assuming that the static friction force is at its maximum, and then solving.
The following forces are at work on the ladder: normal force from wall F up, normal force from floor uF to the right, static friction force from floor, mg down, ladder weight P down, human weight W to left (equal to the coeff times the normal)
The result of adding the horiz and vert forces is:
horiz: $uF=W$
vert: $mg+P=F$
The torques, too? Pick a rotational axis. Given that the ladder is stationary, you can select any position. The place where the ladder meets the ground is the one I would pick.
Every force now generates torque equal to rFsin? where? is the angle between r and F, r is the distance between the point of rotation and the place where the force is applied, and F is the force. For uF and F in our situation, r is zero since they are applied at the selected point of rotation. Regarding the other forces:
mg r is $\frac{L}{2}(4m)$ because the force of gravity acts atthe center of the ladder
P because the individual is at this distance from the ladder's bottom.
The angle "r" creates with the horizontal forces is 50 degrees, whereas it makes an angle of 40 degrees with the vertical forces, hence W r is L (8 m) since this force operates at the other end of the ladder because the ladder is 50 degrees from the horizontal. So, this is the torque sum:
$mg(\frac{L}{2})\sin 40+Px\sin 40=WL\sin 50$
So here's your three equations:
$mg(\frac{L}{2})\sin 40+Px\sin 40=WL\sin 50$
$uF=W$
$mg+P=F$ andyour three unknowns are x, W, F
Fill in the middle one after solving the last one for F. To remove W, substitute that one into the first one. You are left with:
$mg(\frac{L}{2})\sin 40+Px\sin 40=u(mg+P)L\sin 50$

At this moment, the following numbers can be entered:
${\displaystyle 200\times \left(\frac{8}{2}\right)\times \sin 40+800\times x\times \sin 40=0.6(200+800)\times 8\times \sin 50}$
Crunch the numbers and:
$514.23+514.23x=3677$
$x=6.15$ meters