Lean a rod against a wall, its upper end on the wall and its lower end on the floor. The first job, before anything else, is to count every force acting on the rod and settle what the problem needs.
The weight is one. It acts at the middle of the rod, because a rod of uniform thickness has its centre of gravity exactly there.
The rest come from the places it touches. From the floor there are two, the normal force perpendicular to the surface and the friction along it. From the wall there is one, the normal force perpendicular to the surface.
The wall gives only one because we have declared it smooth. Smooth means frictionless, so nothing but a perpendicular force can come from it. The two arrows at the floor and the single arrow at the wall in the figure show exactly that difference. Four in all, and that is everything.
With all four in hand, we can write the balance. The vertical direction first.
Only two forces point vertically, the weight and the floor's normal force . The wall's and the floor's friction both point horizontally, so neither enters the vertical account.
They balance, so . Lay the rod down or stand it up in the figure and these two arrows do not stir. Whatever the angle, the floor carries the whole weight of the rod.
That is the first equation. It tells us , but it says nothing about the we want.
The horizontal next. Two forces point that way, the wall's push and the floor's friction .
The wall pushes the rod to the right. Pushed back, the lower end of the rod tries to slip to the right, so the friction holds it and points left. These two balance, giving .
Swing the angle in the figure and the two grow and shrink together. Lay the rod flatter and it presses the wall harder, so the friction comes out stronger to match. Stand it more upright and both grow smaller.
That is the second equation. And here something starts to look wrong.
All that says is that the two are equal. It says nothing whatever about what their value is.
Count them up. The unknowns are , and , three of them. The equations are the vertical and the horizontal, two of them. We are one short. settles , but for and together there is only the single equation left.
In the figure the angle of the rod is held fixed while and are stretched and shrunk together. At any length you please, the vertical balance holds and the horizontal balance holds. The balance of forces has no power to pin these two down.
The missing equation was prepared in the last lecture. Balance of a rigid body takes two conditions, and we have not used the moments yet. Write that condition in the next lecture and is fixed, and the angle of slipping falls out with it.