# A book pressed against a wall, thought analysis.

1. Dec 9, 2011

### Whakataku

Consider that the book and the force applied by the hand is all together one part, so no worries if there is any slippage between the hand and the book. So then, the book is pressed against the wall with some force. The free-body diagram drawn in my head has a force of friction( μ static*normal force), a horizontal normal force, force acted on the book by the gravity, W=m*g. Then I also have a force applied at some angle.

So my first question is does the x-component of the Force applied be the same as the force of normal.

If that is correct, then wouldn't the angle from the wall where it is impossible to hold onto the book is when the Force of normal, or x-component of the force applied is less than or equal to zero. But then, that doesn't right, I'm guessing there is something to do with the force of friction.

So is my thought process right or wrong?

2. Dec 9, 2011

### dacruick

I'm not sure i completely understand your whole post to be honest. The book is not moving in the x-direction or y-direction, therefore the net force has to be 0. So the force that you are pushing on the book with is the same as the force that the wall is pushing back with.

The downwards force of gravity on the book is the same as the upwards force of friction on the book.

3. Dec 9, 2011

### Whakataku

Dacruick you are right, I'm imagining the book to be static. Sorry, I forgot to mention that. The net force is zero, but my question is on the vector component of the forces(specifically the relationship between the x-component of force applied and the normal force), and the angle of the force applied. Again to be specific, the question is how must the forces be in order for the book to fall, such as the force of normal going zero, or something along the line.

Last edited: Dec 9, 2011
4. Dec 9, 2011

### dacruick

if you change the angle of the applied force, the normal force coming from the wall will be equal to the x component of that applied force. the y-component of the applied force will be compensated by the friction force.

If the y component of the applied force exceeds the force of friction, you will have movement.

If you are looking for the mathematical relationship, it will be the applied force multiplied by the sine and cosine of the angle in which the force is being applied.

I'm still not sure if I've answered your question properly, so let me know if I've been unclear somewhere.

5. Dec 9, 2011

### Whakataku

Thank you very much Dacruick for clearing that up. So my thought of the normal force being equal to the x-component of the force applied was the correct assumption.

Now the comment on the y-component of the force applied, was beneficial. You said "If the y component of the applied force exceeds the force of friction, you will have movement." I want a force that will make the book fall. So then considering up as positive and down as negative, when the angle exceeds 90 degrees (relative to the horizontal wall), the y-component of force applied will be the opposite sign of the force of friction. So then is it right to say that when the y-component of the force with the help of force of gravity exceeds the force of friction, then the book will fall.
So basically what I thinking is when sum of all the y-force (W=mg, Fy= Fapp *sinɵ, and Force of friction), and if all of it adds up to negative the book will fall.
Am I right?

Last edited: Dec 9, 2011
6. Dec 12, 2011

### dacruick

Sorry for the late response, I was away all weekend. yes you are correct for the most part.

The friction force will only oppose motion, it cannot create motion. So the friction force can act in either direction.