I got this problem in the book, Elements of Statics and Dynamics. Please help me in the problem! Thanks!

If the resultant of two forces acting on a particle be at right angles to one of them, and its magnitude is one third of the other, show that the ratio of the longer force to the smaller is $3:2\sqrt { 2 }$ Note by Swapnil Das
5 years, 11 months ago

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Let us assume that the resultant force is in the horizontal direction and the force $F_1$ in the vertical direction. The direction of $F_2$ will be as shown in the figure. The vertical component of $F_2$ should be equal to $F_1$ since there is no resultant force in vertical direction. It is given that $F_2 = F_r/3$ i.e. the horizontal component of $F_2 = F_2/3$ . By simple trigonometry we get that $\sin \theta = 1/3$ where $\theta$ is the angle between the vertical and $F_2$ . and therefore, $\cos \theta = 2\sqrt{2}/3$ . Therefore, again by trigonometry, $F_2 = F_1 \times 2\sqrt{2}/3$ Therefore, we get the required ratio !! i'm not able to post a figure now, will post it later.

- 5 years, 11 months ago

If we assume the resultant of the given forces to lie on the co-ordinate axes such that the resultant is along the x-axis, F1 is along the y-axis and F2 is in the third quadrant making an angle A with the x-axis, then this problem is a breeze... component of F2 along y-axis is equal to F1, since the resultant is perpendicular to it,and only the component of F2 along the resultant(x axis), has contributed to the resultant..... HOPE IT HELPS....

- 5 years, 11 months ago

Draw them as a right triangle. Use geometry for it. Take the resultant as $x$. The other vector will be $3x$. Now take the ratio.

- 5 years, 11 months ago

Can you make a effort to show the process?

- 5 years, 11 months ago

Dude its just pythagoras theorem.

- 5 years, 11 months ago

Thank You brother, I got your method!

- 5 years, 11 months ago

Any day!

- 5 years, 11 months ago

Oh, I am getting a physics feeling...

- 5 years, 11 months ago