- #1
Lone Wolf
- 10
- 1
- Homework Statement
- A 15.0-m 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 800-N 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.