# Book against a wall and forces

If you take a book with mass of 1kg and push it against the wall. With how much force do you have to push the book so it does not fall?

The problem is I know how to calculate this problem, you say $F_{friction}$=$F_{gravitational}$ and $F_{wall / normal}$ = $F_{human / push}$.

The problem is solvable if you say that $F_{gravitational}$ = $F_{normal / wall}$, but why is this true?

How would you calculate this problem if you didn't know that $F_{normal}$ = $F_{gravitational}$, how would you prove this statement?

For me the problem is that $F_{normal}$ = $F_{gravitational}$ * $\cos(\alpha)$, but cos(angle) is 0, I don't understand the relationship between $F_{gravitational}$ and $F_{normal}$ in this scenario.

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You may find it helpful to use LaTeX markup in you posts to Physics.SE (enabled by the MathJax rendering engine). There are some minimal examples in the faq, and you will be interested in a form like F_{gravitational} placed between single dollar signs to get $F_{gravitational}$. Slightly better is F_\text{gravitational} ($F_\text{gravitational}$) if you care. Use \sin \alpha and \cos (\theta) to get the correct markup for standard functions ($\sin \alpha$, $\cos (\theta)$) – dmckee Nov 22 '11 at 1:29
Now, the force exerted by the wall on the book, breaks into two parts (by definition): the normal force which is always perpendicular to the wall (so for a vertical wall can't counteract gravity), and a possible transverse force at the surface. What categories of forces do you know that act at the interface between two solids? What functional relationships exist between them and other forces in the problem? – dmckee Nov 22 '11 at 1:38
Why should the normal force equal the gravitational force? Experiment: Grease up your wall, but an iron plate into the book and a strong electromagnet on the other side (to rule out the friction on your hand). You can be sure the book will slide down the wall not only when the magnetic force is 10N, but for much higher values. Perhaps I don't understand the question ... – mcandril Nov 25 '11 at 13:24

What you need is a relationship between the frictional force and the normal force.

The Wikipedia article on friction has $$F_\mathrm{f} \leq \mu F_\mathrm{n}$$ where $\mu$ is the coefficient of friction. You want to find the minimum normal force necessary, i.e. when this inequality becomes an equality.

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I think it would help if you rotate the figure by 90degrees. Instead of seeing it as what force to push the book with, make the book horizontal. Like this:-

So, now you push the book downwards, the force due to gravity will act leftwards, and the fictional force will be rightwards.

Now the question becomes simpler. From the figure, assume that $F_{g}$ is a force acting on the book to the left. $F_{f}$ is the frictional force and $F_{push}$ is the force you apply(which will act analogous to gravity in actual horizontal systems). The question now converts to what should the frictional coefficient be so the book doesn't move. So we equate $F_{f}$=$F_{g}$. Since,
$F_{g}$ = mg (where 'm' is mass of the book and 'g' is the gravitational constant)

Hence,

$F_{f}$ = mg

or, since in $F_{f}$ is just friction coefficient times your applied force

u$F_{push}$ = mg (where $F_{push}$ is the force you apply)

therefore, $F_{push}$= mg/u .

So as you can see its not a matter of proving the statement $F_{frictional}$=$F_{gravitational}$ , but in order to satisfy the condition that the book doesn't move, $F_{frictional}$=$F_{gravitational}$, becomes a condition which must be satisfied(else the book would move).

Edit:: I tried to add an image to explain the figure, but can't do so because of low reputation points.
Edit:: Got enough reps to add images now, hope this clears it up.

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Hi Likhit, and (belatedly) welcome to Physics Stack Exchange! If I'm reading it correctly, your answer explains why frictional force has to equal gravitational force, but the question is about whether and why the normal force equals gravitational force, so you may have been a bit off the mark. Other than that, though, it's a well-prepared answer. I especially wanted to compliment your use of diagrams, which we could always use more of around here :-) – David Z Nov 25 '11 at 8:24