 9
 1
 Problem Statement
 A 15.0m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.0° angle with the horizontal. Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800N firefighter has climbed 4.00 m along the ladder from the bottom.
 Relevant Equations

Conditions for stactic equilibrium:
Στ = 0
and
ΣF = 0
I solved this question correctly, however I have a question regarding how I should work with the weight of the firefighter climbing the ladder. When drawing the force diagram for this problem, I should only include forces acting on the ladder, right? Which means I would represent the normal reaction force the firefighter makes on the ladder and not the weight of the firefighter  but the textbook shows the weight of the firefighter.
For the condition of rotational equilibrium this distinction doesn't matter much since the result would be the same:
Στ = 0 (around the contact point with the ground):  Wfirefighter*sin(30°)*4  Wladder*sin(30°)*7.5 + nwall*sin(60°)*15 = 0, nwall = 267.5 N
 nfirefighter*sin(90°)*4  Wladder*sin(30°)*7.5 + n*sin(60°)*15 = 0, since nfirefighter = Pfirefighter cos(60°).
However the I found the result for the second part of the problem (finding the reaction force of the ground and the friction force) to be different depending on whether I use the normal force of the firefighter or the weight of the firefighter.
If I use the weight of the firefighter (this is the textbook's solution):
ΣFy = 0: nground  Wladder  Wfirefighter = 0, nground = 1300 N (upwards)
ΣFx = 0: nwall = fa, fa = 267.5 N (opposite direction as the reaction force of the wall).
If I use the normal force of the firefighter on the ladder it must be decomposed in the x and y components.
ΣFy = 0: nground  Wladder  nfirefighter*cos(60°) = 0, nground = 500 + 800*cos²(60°) = 700 N (upwards)
ΣFx = 0: fa  nwall + nfirefighter*sin(60°) = 0, fa =  78,9 N (same direction as the reaction force of the wall).
I don't understand why it's incorrect to use the reaction force of the firefighter instead of the weight of the firefighter. If anyone can clarify that for me I would really appreciate it.
For the condition of rotational equilibrium this distinction doesn't matter much since the result would be the same:
Στ = 0 (around the contact point with the ground):  Wfirefighter*sin(30°)*4  Wladder*sin(30°)*7.5 + nwall*sin(60°)*15 = 0, nwall = 267.5 N
 nfirefighter*sin(90°)*4  Wladder*sin(30°)*7.5 + n*sin(60°)*15 = 0, since nfirefighter = Pfirefighter cos(60°).
However the I found the result for the second part of the problem (finding the reaction force of the ground and the friction force) to be different depending on whether I use the normal force of the firefighter or the weight of the firefighter.
If I use the weight of the firefighter (this is the textbook's solution):
ΣFy = 0: nground  Wladder  Wfirefighter = 0, nground = 1300 N (upwards)
ΣFx = 0: nwall = fa, fa = 267.5 N (opposite direction as the reaction force of the wall).
If I use the normal force of the firefighter on the ladder it must be decomposed in the x and y components.
ΣFy = 0: nground  Wladder  nfirefighter*cos(60°) = 0, nground = 500 + 800*cos²(60°) = 700 N (upwards)
ΣFx = 0: fa  nwall + nfirefighter*sin(60°) = 0, fa =  78,9 N (same direction as the reaction force of the wall).
I don't understand why it's incorrect to use the reaction force of the firefighter instead of the weight of the firefighter. If anyone can clarify that for me I would really appreciate it.