# Physics Problem #1



## ShadenSmith (Aug 22, 2008)

An 8.00m, 200 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.600, and the ladder makes a 50.0 degrees angle with the ground. How far up the ladder can an 800 N person climb before the ladder begins to slip?


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## JBCM627 (Aug 22, 2008)

This looks familiar...


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## badmephisto (Aug 22, 2008)

argh, i solved this problem in my first year physics, and one thing i remember about it is that it was not easy. There are torques involved  I'm scared of torques


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## ShadenSmith (Aug 22, 2008)

You're quite right. Torques are involved


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## rachmaninovian (Aug 22, 2008)

i'm doing it in american equivalent of grade 9 >.<
I hate torques...scared of it...


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## ShadenSmith (Aug 22, 2008)

Oh, they're not that bad.


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## Rosson91 (Aug 22, 2008)

spoiler below

my result is 5 m.


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## Rosson91 (Aug 22, 2008)

I think I cant explain that in English....if they say it is correct I'll try to explain it


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## ShadenSmith (Aug 22, 2008)

Rosson91 said:


> I think I cant explain that in English....if they say it is correct I'll try to explain it




Well...watch your rounding. Try to give me an answer that's rounded to two decimals, and I'll tell you. You may or may not be correct. It's in that ballpark though.


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## Odin (Aug 22, 2008)

thats cool how you did the spoiler warning


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## Rosson91 (Aug 22, 2008)

spoiler below
4.99 m


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## ShadenSmith (Aug 22, 2008)

Rosson91 said:


> spoiler below
> 4.99 m



Nope. (too short)


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## not_kevin (Aug 22, 2008)

By the way, using spoiler tags ([ spoiler ] ... [ /spoiler ] without the extra spaces) around text you want to hide is better; you can still sometimes see white text (plus it's easier for you).


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## hawkmp4 (Aug 22, 2008)

Ugh...
I had the WORST teacher for AP physics last year. We never even discussed fluids, torque, or modern physics. I'm lost on this one.


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## not_kevin (Aug 22, 2008)

hawkmp4 said:


> Ugh...
> I had the WORST teacher for AP physics last year. We never even discussed fluids, torque, or modern physics. I'm lost on this one.



Sounds like mine! Mechanics only; plus, mine never even covered SHM. The only reason I got a 5 was because I knew a little math to help me.


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## Rosson91 (Aug 23, 2008)

I tried to solve it again
spoiler below
6.22 m


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## ShadenSmith (Aug 23, 2008)

Rosson91 said:


> I tried to solve it again
> spoiler below
> 6.22 m




I think that's probably close enough to say you got it right.


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## JBCM627 (Aug 23, 2008)

ShadenSmith said:


> Rosson91 said:
> 
> 
> > I tried to solve it again
> ...



@Rosson, please use spoiler tags.

I guess I'll post a solution (although I will say I did already know the answer to this before it was posted... and none of my initial attempts to solve this were right):


Spoiler



Setup: The person is an arbitrary distance up the ladder of length L, say, x. The ladder's mass will be m, and the person's mass will be M.

Picking the base of the ladder as an origin is nice, as we don't have to consider the torque generated by the base of the ladder. 3 more objects will apply a force to the ladder: gravity and the person in a downwards direction, and the wall in a horizontal direction.

Torque is *r* x *F*, or the cross product between the vectors F and r. F is force, and r is radius (distance from your origin). In a more basic sense, it is r*F, where F is the componet of *F* perpendicular to *r*, and r is |*r*|.
Drawing a picture to see this helps, but:
The torque from the ladder will be (L/2)*mg*cos(50).
The torque from the person will be x*Mg*cos(50).
The torque from the wall will be L*Fw*sin(50), where Fw is the force exerted by the wall.

Balancing these, we have:
((L/2)*m + x*M)g*cos(50) = L*Fw*sin(50).

We want to find when the ladder starts to slip, or when Fw = mu*Fn = mu*(M+m)g.

Plugging in and simplifying we have:
((L/2)*m + x*M)cos(50) = L*mu*(M+m)*sin(50).

Plugging in and solving for x, we have:
x = (L*mu*(M+m)*tan(50) - (L/2)*m)/M
= (8*600*tan - 800)/800
= 6*tan(50) - 1
= 6.15


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