Man on a Ladder

## **Problem**

**Part 1:** A ladder of mass

m

{\displaystyle m}

 and length

l

{\displaystyle l\;}

 leans aginst a smooth building and has a coefficient of static friction

μ
s

{\displaystyle \mu\_ s}

 with the ground. A man of mass

M

{\displaystyle M}

 stands a distance

x

{\displaystyle x\;}

 up the ladder. At what angle

θ

{\displaystyle \theta\; }

 with the ground will the ladder begin to slip? Express the result in terms of

tan
⁡
θ

{\displaystyle \tan \theta }

.

**Part 2:** Suppose the mass of the ladder is 20 kg, the length of the ladder is 5 m, the mass of the man is 80 kg, and the coefficient of static friction is 0.30. If the ladder makes and angle of 60 degrees with the ground, how far up the ladder can the man climb before the ladder starts to slip?

## **Solution**

**Part 1**

Draw a free-body diagram and label all the forces. Let

F
w

{\displaystyle F\_w}

 be the force from the wall on the ladder,

F
f

{\displaystyle F\_f}

 be the force of friction, and

F
N

{\displaystyle F\_N}

 be the normal force. The weight of the man

M
g

{\displaystyle Mg }

 is located some distance

x

{\displaystyle x\;}

from the base of the ladder. The center of gravity of the ladder

m
g

{\displaystyle mg}

 is located at the middle of the ladder

l

/

2

{\displaystyle l/2}

.

Since this configuration is static, the net accelerations in both the vertical and horizontal directions are zero. Write the net forces in the vertical and horizontal directions:

F
N

−
m
g
−
M
g

=
0

F
w

−

F
f

=

{\displaystyle \begin{align}
F\_N - mg - Mg &= 0 \\[5 pt]
F\_w - F\_f &= 0.
\end{align}}

This shows

F
N

=
(
m
M
)
g

F
w

=

F
f

.

{\displaystyle \begin{align}
F\_N &= (m+M)g \\[5 pt]
F\_w &= F\_f.
\end{align}}

There are three unknown forces but only two equations, so make a third equation using torque. The net torque is zero because the ladder is stationary; therefore, the clockwise and counterclockwise torques are equal in magnitude. Designate the point where the ladder touches the floor as the pivot point, thus eliminating

F
f

{\displaystyle F\_f}

 and

F
N

{\displaystyle F\_N}

 in the torque equation. The force

F
w

{\displaystyle F\_w}

 contributes to the clockwise torque; the two forces of gravity

M
g
,

m
g

{\displaystyle Mg, \; mg }

 contributes to the counterclockwise torque.

τ

c
w

=

τ

c
c
w

F
w

(
l
)
sin
⁡
θ

=
m
g
(
l

/

2
)
sin
⁡
(

90

∘

−
θ
)
M
g
(
x
)
sin
⁡
(

90

∘

−
θ
)

F
w

(
l
)
sin
⁡
θ

=
m
g
(
l

/

2
)
cos
⁡
θ
M
g
(
x
)
cos
⁡
θ

{\displaystyle \begin{align}
\tau\_{cw} &= \tau\_{ccw} \\[5 pt]
F\_w (l) \sin\theta &= mg(l/2)\sin(90^{\circ} - \theta) + Mg(x)\sin(90^{\circ} - \theta) \\[5 pt]
F\_w (l) \sin\theta &= mg(l/2)\cos\theta + Mg(x)\cos\theta
\end{align}}

Remember that

sin
⁡
(

90

∘

−
θ
# )
cos
⁡
θ

{\displaystyle \sin(90^{\circ} - \theta) = \cos\theta}

. Solving for

F
w

{\displaystyle F\_w}

 gives

F
w

=

(

1
2

m
l
M
x
)
g

l

cos
⁡
θ

sin
⁡
θ

=

(
m
l
2
M
x
)
g

2
l
tan
⁡
θ

.

{\displaystyle \begin{align}
F\_w &= \frac{(\frac{1}{2}ml + Mx) g}{l} \frac{\cos\theta}{\sin\theta} \\[5 pt]
&= \frac{(ml + 2Mx)g}{2l\tan \theta}.
\end{align}}

Solving for

F
w

{\displaystyle F\_w}

 also solves for the magnitude of

F
f

{\displaystyle F\_f}

:

F
f

=

(
m
l
2
M
x
)
g

2
l
tan
⁡
θ

.

{\displaystyle F\_f = \frac{(ml + 2Mx)g}{2l\tan \theta}. }

Because the ladder is not sliding yet, the static friction formula

F
f

≤

μ
s

F
N

{\displaystyle F\_f \le \mu\_s F\_N }

 shows that

(
m
l
2
M
x
)
g

2
l
tan
⁡
θ

≤

μ
s

(
m
M
)
g
.

{\displaystyle \frac{(ml + 2Mx)g}{2l\tan \theta} \le \mu\_s (m+M)g. }

Solving for

tan
⁡
θ

{\displaystyle \tan \theta }

 gives

tan
⁡
θ
≥

m
l
2
M
x

2

μ
s

(
m
M
)
l

.

{\displaystyle \tan \theta \ge \frac{ml + 2Mx}{2\mu\_s (m+M)l}. }

**Part 2**

Use the result from part 1 and substitute the given numerical values. Make sure to change the inequality into an equality. The numerical values are

# m
20

k
g
,

# M
80

k
g
,

μ
s

=
0.3
,

# l
5

m
.

{\displaystyle m = 20\; kg,\; M = 80 \; kg,\; \mu\_s = 0.3,\; l = 5\; m. }

tan
⁡
(

60

∘

)

=

20
(
5

m
)
2
(
80

k
g
)
x

2
(
0.3
)
(
20

k
g
80

k
g
)
(
5

m
)

tan
⁡
(

60

∘

)

=

100
160
x

300

x

=
2.62

m

{\displaystyle \begin{align}
\tan(60^{\circ}) &= \frac{20(5 \; m) + 2(80 \; kg)x}{2(0.3)(20 \; kg + 80\; kg)(5 \; m)} \\[5 pt]
\tan(60^{\circ}) &= \frac{100 + 160x}{300} \\[5 pt]
x &= 2.62 \;m
\end{align} }

# x
2.62

m

{\displaystyle x = 2.62 \; m }