Find Coefficient of Friction for Firefighter on Ladder

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To find the coefficient of friction for a firefighter on a ladder leaning against a frictionless wall, the problem involves analyzing forces and torques using Newton's laws. A free-body diagram is essential to visualize the forces acting on the ladder, including the weight of the firefighter and the ladder itself. The angle of the ladder can be determined using trigonometry, specifically tan^-1(4/3), yielding approximately 53.1 degrees. The relationship between the angle and the coefficient of friction is expressed as tan(53.1) = 2u, leading to a calculated coefficient of friction of approximately 0.67. This approach effectively simplifies the problem despite the complexity of the variables involved.
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Homework Statement


(8 marks)
A uniform ladder (m = 35kg) leans against a smooth frictionless wall. The foot of the ladder is 3 m from the wall and an 85kg firefighter stands 3m up the ladder as shown. Find the coefficient of friction of the floor.
http://i49.tinypic.com/ftdo9g.jpg

Homework Equations


(Angular Newton's Second Law and Linear Newton's First Law)
Στ=0 and ΣFx=ΣFy=0
τ = F⊥r

The Attempt at a Solution


 
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Draw a free-body diagram of the ladder, being sure to include all forces. Then write out Newton's second law for both the horizontal and the vertical directions. For torque, choose a point and write out the rotational analogue of Newton's second law about that point.
 
i've tried that, and it gets so complicated and there are so many unknown variables that I get lost every single time. this was actually on my test and i got 0.
 
ladder is 3m from wall...it connects with the wall 4m high...so get the angle which is tan^-1 of 4/3...which is 53.1

then tan53.1=2umg/mg...which is equal to 2u..
tan53.1/2=coefficient friction=.67

i think that's how you do it
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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