Can a Painter Safely Stand on an Angled Ladder?

  • Thread starter Thread starter blue2004STi
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 3K views
blue2004STi
Messages
21
Reaction score
0

Homework Statement


A painter wishes to know whether or not she can safely stand on a ladder. The ladder has a mass M1 = 14 kg which is uniformly distributed throughout its length L = 8.6 m. The ladder is propped up at an angle theta = 52o. The coefficient of static friction between the ground and the ladder is mus = 0.45, and the wall against which the ladder is resting is frictionless. Calculate the maximum mass of the painter for which the ladder will remain stable when she climbs a distance d = 5.2 m up the ladder. (The painter's mass might be so low that only Lilliputian painters can safely ascend the ladder.)



Homework Equations


Fric = Us*N
F=m*a
Torque= r *(dot)F

The Attempt at a Solution



Right now I'm just looking for a poke in the right direction. I can't even figure out where to start. I know that the frictional force is going to be at a maximum when the weight is at the maximum. I think that the normal force will be the 137.2 N from the ladder + 9.8*the mass of the painter. But how do I figure out what the maximum Fric will be? I also know that the problem has to do with torque. Any help is appreciated, I'm new so I'm sorry if I didn't give it enough effort but I've been staring at it and my book for about an hour to an hour and a half and couldn't get anywhere.


Thanks,
Matt
 
Physics news on Phys.org
First step draw a free body diagram and list every single force that applies. The floor puts a normal force on the ladder, the wall puts a normal force on the ladder, weight of the ladder, weight of the painter. This is a statics problem meaning that the sum of the forces need to equal 0, if they didn't then she'd be falling. Try summing forces in the x and y directions. Once you find that you have too many unknowns try summing torques to your advantage, they have to sum to 0 as well. I hope that starts you in the right direction.
 
So I drew it out after dinner and I came up with this.
If m=mass of the painter, and Nx is the force by the ladder on the painter(in the x-direction), and Ny is the force by the ladder on the painter(in the y-direction) then my equations look like:

(9.8m + 14*9.8)Cos(52) = .45(9.8m + 137.2)N - In the x-direction

Ny = (9.8m + 137.2) - y-direction

Ny^2 + Nx^2 = N^2

I realize these are probably off and you all are chuckling at home at my attempt :smile: but any help is appreciated. Still stuck.
Thanks,
Matt