Ladder Equilibrium: Can 65kg Person Climb Without Slipping?

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SUMMARY

The discussion centers on a physics problem involving a uniform 5.0-kg ladder leaning against a frictionless wall at a 15-degree angle, with a coefficient of friction of 0.26 between the ladder and the ground. A 65-kg person is attempting to climb the ladder without it slipping. The key equations used include torque and force equilibrium, specifically the equation involving torque: F_W L sin(90° - 15°) - M_p gL sin(15°) - M_L g(L/2) sin(15°) = 0. The correct interpretation of angles is crucial, as the angle with the floor is 75 degrees, not 15 degrees.

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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


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

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:

<br /> 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<br />

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!
 
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Hi RedBarchetta,


RedBarchetta said:

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:

<br /> 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<br />

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?
 

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