Calculate the force on a circular surface

[jonjacson]

Hi to everybody

Homework Statement

(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.

Homework Equations

Newton laws and basic trigonometric relationships.

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

 

[lewando]

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

 

[TSny]

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

And I can calculate A as:

A=arcsen(s/r)

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.

 

[TSny]

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

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

 

[jonjacson]

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.

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

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.