Homework Help: Calculate the force on a circular surface

1. Dec 28, 2012

jonjacson

Hi to everybody

1. The problem statement, all variables and given/known data

(The attachment has an image with this problem)

A small probe P is gently forced against the circular surface with a vertical force F as shown. Determine the n- and t-components of this force as functions of the horizontal position s.

2. Relevant equations

Newton laws and basic trigonometric relationships.

3. The attempt at a solution

Well I show in the attachment my reasoning, the force F is applied at the contact point C, I have represented the direction in which this force acts (vertical line). I need to find the proyections of F to the n and t axis, to calculate Fn and Ft in red.

Well I think that the angle between the vertical line and the line connecting the center O and the point C is the same angle between F and Fn, in the picture they are called A. I simply write:

Fn= F cos(A)
Ft= F sen(A)

And I can calculate A as:

A=arcsen(s/r)

But I have seen the answers provided in the book and they are:

Ft=Fs/r

and

Fn= -(F√(r2-s2))/r

Does anybody know what is Fs?

And I have no idea where the equation for Fn comes from.

Why are my answers so different to the ones in the book?

Thank you

Attached Files:

• 1111111.png
File size:
7.1 KB
Views:
354
2. Dec 28, 2012

lewando

"Fs" is F*s, not Fs. Looks like they are avoiding using sin, cos in the answer.

3. Dec 28, 2012

TSny

Looks good. (The sine function is generally written sin rather than sen.) From A=arcsin(s/r) you get sin(A) = s/r. Try substituting this expression for sin(A) into your expression Ft= F sin(A) and compare to answer in book.

4. Dec 28, 2012

TSny

How would you write cos(A) in terms of s and r?

5. Dec 28, 2012

jonjacson

Yes you are right, thank you very much.

The sinus is simply s/r and I get the Ft.

For the Fn it´s the same, they avoid to use the cos, to calculate the other leg of the triangle I use the pythagorean theorem r2=s2+y2 where I called y the other leg of the triangle. So the cosinus is exactly the answer in the book.

Thank you very much to both of you, I have more doubts with problems in this book, hope to see you in the other threads.