Ladder Physics Problem?
A 10-kg ladder leans against a (frictionless) wall at 50 degrees. An 80-kg man climbs 1/3 of the way up the ladder. What is the magnitude of the contact force with the wall?
Answer: 260 N
At first glance, this problem seemed easy to me, but then my mind went blank and it really confused me. Please show the solution if you know how. Thank you so much!
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|.--------------------> N₁
|..\
|50°.\
|.........\
|............\
|...............\
|...............|..\
|...............|.....\..........N₂
|...............|........\.......⋀
|..............V........|..\.....|
|.............mg. .....|.....\..|
|______________|.fr <--\|______________________
.........................|
........................V
.......................Mg
Having drawn all the acting forces, you just have to balance torques(since the ladder is in equilibrium)
On balancing torques about lower end of the ladder,
mg*L/2 Sin50 + Mg*L/3 Sin50 - N₁*L Cos50 = 0
N₁ = [ mg/2 + Mg/3 ] Sin50/ Cos50
N₁ = [ mg/2 + Mg/3 ] *Cot50
N₁ = [49 + 261.33 ] * 0.839
N₁ = 260.36 N
Yes, the "ladder problem" can be a little tricky until you work one out.
I did this for an earlier problem this year (ref. 2); here's the basic approach (I'll quote from it):
"
Approach:
Analyze the forces and moments on the system just before the ladder begins to slip. At this point in time the forces and torques must be in balance, since nothing is moving. Then use the definition of μs, the coefficient of static friction, to calculate it. There is an illustration with some limited animation at reference 1; I'll use the same notation, even if it seems non-standard.
At the base of the ladder, there are two forces operating:
An upward force normal to the floor Fn supporting the ladder; and
a horizontal frictional force H that keeps the ladder from slipping.
At the top of the ladder, there is a force F normal to the wall to keep the ladder from moving.
. . . "
I hope this can get you started on it!
.