Ladder Equilibrium: Can 65kg Person Climb Without Slipping?

[RedBarchetta]

Homework Statement

http://img206.imageshack.us/img206/8178/ladderjt1.png A uniform 5.0-kg ladder is leaning against a frictionless vertical wall, with which it makes a 15* angle. The coefficient of friction between ladder and ground is 0.26. Can a 65-kg person climb to the top of the ladder without it slipping? If not, how high can the person climb? If so, how massive a person would make the ladder slip?

Homework Equations

$$\begin{gathered}<br /> \sum {\vec \tau } = \vec 0 \hfill \\<br /> \sum {\vec F} = \vec 0 \hfill \\ <br /> \end{gathered}$$

The Attempt at a Solution

So, I choose the end of ladder that is touching the floor to be the axis. Now I want to find the sum of the torques, then set them equal to zero. What I don't quite get in the equation is the sin(90-15) in the first term in the following equation my professor gave me:

$$F_W L\sin (90^ \circ - 15^ \circ ) - M_p gL\sin (15^ \circ ) - M_L g\frac{L}<br /> {2}\sin (15^ \circ ) = 0$$

sin(90-15)? Technically, it should be the angle between the r and F...but this doesn't make sense to me...or perhaps 15* is meant to be the other angle in the triangle?

If anyone can help, it would be greatly appreciated. Thank you!

 

[alphysicist]

Hi RedBarchetta,

The Attempt at a Solution

So, I choose the end of ladder that is touching the floor to be the axis. Now I want to find the sum of the torques, then set them equal to zero. What I don't quite get in the equation is the sin(90-15) in the first term in the following equation my professor gave me:

$$F_W L\sin (90^ \circ - 15^ \circ ) - M_p gL\sin (15^ \circ ) - M_L g\frac{L}<br /> {2}\sin (15^ \circ ) = 0$$

sin(90-15)? Technically, it should be the angle between the r and F...but this doesn't make sense to me...or perhaps 15* is meant to be the other angle in the triangle?

Yes, it's the other angle; the ladder makes a 15 degree angle with the wall, so the angle with the floor is 75 degrees. Once you correct your diagram, do you see why the angles your professor chose are correct?