# A STEP question that just drops out

If there’s one thing I like better than a STEP mechanics question, it’s a STEP mechanics question that just drops out when you get the diagram right.

It’s question 11 from the 2017 STEP I: a ladder in a box on a slope.

I’m going to answer it twice, once using the “routine” diagram-forces-moments-algebra method, and then the really nice way. But first, let’s get the diagram right.

### The mandatory diagram improvement

One of the best hacks I know for awkward inclined plane diagrams is to tilt the page so the floor is flat and the horizontal and vertical are on an incline. Sure, this means you have to account for the weight not acting vertically, but (in this case) it means all of the other forces are vertical and horizontal.

### The three equations

Coming up with three equations is fairly straightforward:

- Resolving parallel to the plane: $R_B = \mu R_A + W \sin(\alpha)$ (1)
- Resolving perpendicular: $R_A + \mu R_B = W \cos(\alpha)$ (2)
- Taking moments about the midpoint: $L\mu R_B \sin(\beta) + L R_B \cos(\beta) + L \mu R_A \sin(\beta) = L R_A \cos(\beta)$ (3)

Now we can work on eliminating as many variables as possible. One is very easy to remove: there’s a common factor of $L$ through the last equation, so we can divide that out:

- $\mu R_B \sin(\beta) + R_B \cos(\beta) + \mu R_A \cos(\beta) = R_A \sin(\beta)$ (3’)

I’m also going to multiply equation (2) by $\tan(A)$ so I can eliminate $W\sin(\alpha)$:

- $R_A \tan(\alpha) + \mu R_B \tan(\alpha) = W \sin(\alpha)$ (2’)

So

- $R_B = \mu R_A + R_A \tan(\alpha) + \mu R_B \tan(\alpha)$ (1’)

Or, if $t = \tan(\alpha)$:

- $R_B(1 - t\mu) = R_A(\mu + t)$ (1’’)

We can treat equation (3’) similarly, letting $T = \tan(\beta)$:

- $R_B(1 + \mu T) = R_A(T-\mu)$ (3’’)

Dividing these last two gives:

- $\frac{1-t\mu}{1+\mu T} = \frac{\mu + t}{T-\mu}$

Which becomes:

- $(1 - t\mu)(T - \mu) = (\mu + t)(1 + \mu T)$

This algebra is already tedious, isn’t it?

- $T - \mu - tT\mu + t\mu^2 = \mu + t + \mu^2 T + \mu Tt$
- $(T-t) - (T-t)\mu^2 = 2\mu + 2\mu Tt$
- $(T-t)(1-\mu^2) = 2\mu(1 + Tt)$

Now cross-divide:

- $\frac{T-t}{1+Tt} = \frac{2\mu}{1-\mu^2}$

Of course, $T$, $t$ and $\mu$ are all $tan$s in disguise:

- $\frac{\tan(\beta) - \tan(\alpha)}{1 + \tan(\alpha)\tan{\beta}} = \frac{2\tan(\gamma)}{1 - \tan^2(\gamma)}$

And those are compound angle identities!

$\tan(\beta - \alpha) = \tan(2\gamma)$.

We know that $\piby 2 > \beta - \alpha > 0$ and that $0 < 2\gamma < \pi$, so the only option is for $\beta-\alpha = 2\gamma$ as required.

Can you see why I looked for a nicer solution?

### A much nicer method

I learned this trick from @JusSumChick, and am always eager to apply it!

First, I’m going to do something with the contact forces: I can combine the normal reaction force and friction force at $A$ into a single force $C_A$ which acts at an angle of $\gamma$ to the left of the ‘vertical’. Similarly at $B$, the forces can combine into a single force $C_B$ at the same angle $\gamma$ above the ‘horizontal’.

Including $W$, we now have three forces. According to the Jus Sum Trick, the rod can only be in equilibrium if lines of action of these three forces pass through the same point ((You can take moments about anywhere. Take moments about a point where two lines of actions intersect. The moment of the other force can only be zero if it passes through that point.)).

The lines of action of $C_A$ and $C_B$ meet at a right angle at point $G$, so the lines of action and the rod form a right-angled triangle. $W$ acts through the midpoint $M$ of the rod, which is the circumcentre of the triangle. Therefore, point $G$ is a distance of $L$ from $M$ and - in particular - triangle $MGA$ is isosceles, and angles $M\hat G A$ and $G \hat A M$ are equal.

Angle $M\hat G A$ is $\alpha + \gamma$. Angle $G \hat A M$ is $\beta - \gamma$. Therefore $\beta - \alpha = 2\gamma$, as required.

* Edited 2021-07-19 to fix LaTeX