A ladder of length 2L and mass M is positioned on level ground leaning against a wall such that the angle between the ladder and the horizontal is α. The coefficient of static friction between the ladder and the wall and between the ladder and the ground is μstatic =0 .65. The centre of mass of the ladder is halfway along it.
For the ladder to be in mechanical equilibrium:
(i) Write down equations for the total x- andy-components of the 5 forces acting on the ladder.
(ii) Consider torques about the centre of the ladder. In which direction (into or out of the page) does the torque due to each of the 5 forces force act?
(iii) Write down an equation for the sum of the torques about the centre of mass of the ladder.
(iv) Use your equation for the torques to derive an expression for tan α in terms of the magnitudes of the forces acting.
My answers so far:
(appologies for the latex, if someone has time can they just have a quick look and see if there is anything glaringly wrong!)
i)
The sum of the horizontal and vertical forces should equal zero.
Horizontal:
$F_2 = \mu F_1$
Vertical:
$mg = F_1+\mu F_2$
ii)
$\sum_{anticlockwise}=\sum_{clockwise}$
Anti-clockwise:
$\mu F_2\times l\cos \alpha+F_2\times l\sin \alpha+ \mu F_1\times l\sin \alpha$
Clockwise:
$F_1 \times l\cos\alpha$
iii)
In equilibrium, the sum of anticlockwise moments is equal to the sum of clockwise moments at any point.
$\mu F_2\times l\cos\alpha+F_2\times l\sin\alpha+ \mu F_1\times l\sin\alpha = F_1\times l\cos\alpha$
iv)
$(F_2+ \mu F_1)\sin \alpha = (F_1- \mu F_2)\cos\alpha$
$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{(F_1- \mu F_2)}{(F_2+ \mu F_1)}$
$F_2 = \mu F_1$
$ \tan\alpha = \frac{(F_1- \mu^2F_1)}{(2\mu F_1)}$
$(1-\mu^2/2\mu)$
Thank you.
Your answers look mostly correct, but there are a few minor mistakes:
i) The equation for the horizontal component of the forces should be:
$F_2 = \mu \cdot F_1 \cos(\alpha)$
ii) The equation for the clockwise moment should be:
$F_1 \cdot l \sin(\alpha)$
iii) The equation for the sum of torques about the center of mass should be:
$(\mu \cdot F_2 \cdot l \cos(\alpha) + F_2 \cdot l \sin(\alpha) + \mu \cdot F_1 \cdot l \sin(\alpha)) - (F_1 \cdot l \cos(\alpha)) = 0$
iv) Your expression for $\tan(\alpha)$ is correct, but you made a mistake in simplifying it. It should be:
$\tan(\alpha) = \frac{F_1 - \mu^2 \cdot F_1}{F_2 + \mu \cdot F_1}$
Simplifying further, we get:
$\tan(\alpha) = \frac{F_1(1 - \mu^2)}{F_2 + \mu \cdot F_1}$
Therefore, the corrected expression for $\tan(\alpha)$ is:
$\tan(\alpha) = \frac{(1 - \mu^2)}{(1/\mu) + 1}$
Everything else looks good!