Print ViewA Person Standing on a Leaning Ladder

Apr 2008
1
0
Wollongong, Australia
This question was taken from an online quiz. After taken numerous shots at it I'm left dumbfounded. I scanned its complimentary book to no avail.
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What is the minimum coeffecient of static friction
required between the ladder and the ground so that the ladder does not slip?
Express
in terms of m1, m2, d, L, and
.
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topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
This question was taken from an online quiz. After taken numerous shots at it I'm left dumbfounded. I scanned its complimentary book to no avail.
----------------------------------------


What is the minimum coeffecient of static friction
required between the ladder and the ground so that the ladder does not slip?
Express
in terms of m1, m2, d, L, and
.
----------------------------------------
This is a static equilibrium problem, so we need to use the fact that the ladder does not accelerate in either coordinate direction and the fact that the ladder does not rotate about any axis.

What troubles me here is that we have no information about where the weight of the man actually acts. With the information available I am going to have to assume that this weight acts entirely a distance d up from the bottom of the ladder, even though the man's hand on the ladder would seem to indicate otherwise. I am taking \(\displaystyle m_1\) to be the mass of the man and \(\displaystyle m_2\) to be the mass of the ladder. Since the ladder must be assumed to be uniform, this weight acts at a distance L/2 up the ladder.

Let the +x direction be to the right and +y upward.

So Newton's 2nd says:
\(\displaystyle \sum F_x = -f + N_1 = 0\)

\(\displaystyle \sum F_y = N_2 - w_1 - w_2 = 0\)

We don't know f, \(\displaystyle N_1\), nor \(\displaystyle N_2\). Thus we need one more equation to solve the system.

The ladder doesn't rotate about any axis, so we can choose a nice point for ourselves to calculate the torques. I am going to choose the point at the base of the ladder.

\(\displaystyle \sum \tau = w_1 \cdot d ~ sin(90 + \theta) + w_2 \cdot \left ( \frac{L}{2} \right ) ~sin(90 + \theta) - N_1 \cdot L~sin(180 - \theta) = 0\)

I'll leave it to you to solve this system. If you run into troubles, just let me know.

-Dan