Need help finding frictional force in torque problem

[EmptyMerc]

Homework Statement

A ladder having uniform density, a length of 3.4 meters and a weight of 121 N rests against a frictionless vertical wall, making an angle of 75 degrees with the horizontal. The lower end rests on a flat surface, where the coefficient of static friction is Mu = 0.400. A painter having a mass of 80 kg attempts to climb the ladder. How far up the ladder will the painter be when the ladder begins to slip?

Homework Equations

Torque equations and frictional force equations

The Attempt at a Solution

So using the torque equations I got x = [F_wall * 3.4 * sin(75) - 121 * (3.4/2) * sin(165)] / (80 * 9.81 * sin(165)

Where x is the distance from the pivot point and F_wall = Frictional force.

But I think I'm having a problem finding the frictional force.

I know Frictional force = Mu(N) , where N is the normal force of the ground.

So to find N I use the formula 121 + 80(9.81) which equals 905.8 and gives a frictional force of .4(905.8) = 362.32.

But when I plug it in I get x = 5.6 m, which is taller than the ladder and leads me to believe that is not the right way to calculate the frictional force.

So is that the right way to calculate the frictional force, or is there another way?

Help is much appreciated.

 

[Spinnor]

To see the same problem worked out, different numbers, see page 270 of,

http://highered.mcgraw-hill.com/sites/dl/free/0072564369/299123/GRR_2e_ch08.pdf

Found from,

https://www.google.com/webhp?hl=en#...f3e5f3c724b693&bpcl=38093640&biw=1093&bih=464

 

[Simon Bridge]

Do the algebra before putting the numbers in ... helps with troubleshooting.
Otherwise working out what you are doing involves some pain sifting through the numbers.

A ladder of mass $M$ and length $L$ sits on a horizontal floor with friction coefficient $\mu$ leaning at angle $\theta$ (to the horizontal) against a frictionless vertical wall. A painter-being, mass m, climbs the ladder. We need to know how far, x, along the ladder the being can go without slipping.

So the torque about the floor pivot (say) would be $\tau=(mx+ML/2)\cos(\theta)$ for example... and the force down the length of the ladder towards the floor-pivot would be $F=(m+M)g\sin(\theta)$ ... this force has a component at the pivot that is directly down and another that it horizontal away from the wall.

Try reworking your math this way - it should be clearer.

 

[EmptyMerc]

Yea it is a lot clearer. I still come up with the same answer and judging on how how the book did it I think it is correct now.

Really appreciate the help though!

 

[Simon Bridge]

Yep - sometimes reworking a problem can clear up that feeling of uncertainty.
When you are presenting your working to someone else, it helps them understand you if you use the symbolic/algebraic form rather than the absolute/numerical form. It would have been a lot for work for me to figure out if you'd done it right or not so I just tried to get you to do the work instead :)