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Static equilibrium -- interpretation of forces

  • Thread starter Lone Wolf
  • Start date
Problem 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.
 

BvU

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Hi,
.. only include forces acting on the ladder, right?
Right
Which means I would represent the normal reaction force the firefighter makes on the ladder and not the weight of the firefighter
Not right: the man exerts a force on the ladder.
The reaction force (ie normal force) is exerted by the ladder on the man (and compensates the gravitational force on him).
 
Hi,
Right
Not right: the man exerts a force on the ladder.
The reaction force (ie normal force) is exerted by the ladder on the man (and compensates the gravitational force on him).
But according to Newton's third law, wouldn't the force the man exerts on the ladder be equal to the reaction force exerted by the ladder on him?
 

BvU

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Equal and opposite
 

haruspex

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The reaction force (ie normal force) is exerted by the ladder on the man
Seems reasonable to me that if A exerts a normal reaction on B then B exerts an equal and opposite normal reaction on A.
the normal reaction force the firefighter makes on the ladder and not the weight of the firefighter
It's a ladder, not a ramp. If it were a ramp then the normal reaction would be perpendicular to the ramp, but there would also have to be a frictional force.
If you stand on the rung of a ladder, without depending on any friction, what is the orientation of the tangent plane between your foot and the rung?
 

jbriggs444

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If the firefighter is supported in equilibrium by forces from ladder, gravity and nothing else then there will be a force of firefighter on ladder. The labels that one chooses to put on that force, "normal", "reaction" or whatever else are irrelevant. What matters is the direction, magnitude and point of application of the force on the ladder.

Those can all be determined without bothering about labels.
 

CWatters

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.. the normal reaction force the firefighter makes on the ladder and not the weight of the firefighter
That sentence is a bit confused.

The weight of the firefighter acts on the ladder.
The normal force is the force the ladder makes on the firefighter.

The problem statement asks about forces acting on the ladder so it seems sensible to construct a free body diagram for the ladder. As a rule a free body diagram contains just one body (in this case the ladder) and all the forces acting on the body (but none of the forces the body applies to other objects). So in this case you want the weight of the fireman acting on the ladder.
 

haruspex

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The normal force is the force the ladder makes on the firefighter.
It is unclear, but my guess at the issue is that @Lone Wolf thinks the normal force from the ladder would be normal to the line of the ladder, as though it were a ramp. See post #5.
 

CWatters

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Good point. I should have said..

The reaction force is the force the ladder makes on the firefighter.

As in this case the reaction force isn't normal (perpendicular) to the angle of the ladder.
 

haruspex

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Good point. I should have said..

The reaction force is the force the ladder makes on the firefighter.

As in this case the reaction force isn't normal (perpendicular) to the angle of the ladder.
That still is not the point I am making.
The reaction force from the ladder is a "normal" normal force, that is, it is perpendicular to the tangent plane separating the bodies in contact. In this case, those bodies are a rung and a foot. The separating plane will be horizontal.
 

jbriggs444

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The reaction force from the ladder is a "normal" normal force, that is, it is perpendicular to the tangent plane separating the bodies in contact.
It might be. We are not given the details of whether the rungs are round or flattened, whether the ladder is at the correct angle to make the flats horizontal or whether the fireman is tilting his feet up or down and is depending on friction to prevent slippage. In my view, referring to the contact force as "normal" is simply not needed. The net contact force is vertical and is applied at the rung. That is all that we need care about.
 

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