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
3 years, 3 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.

- 3 years, 3 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....

- 3 years, 3 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.

- 3 years, 3 months ago

Can you make a effort to show the process?

- 3 years, 3 months ago

Dude its just pythagoras theorem.

- 3 years, 3 months ago

Thank You brother, I got your method!

- 3 years, 3 months ago

Any day!

- 3 years, 3 months ago

Oh, I am getting a physics feeling...

- 3 years, 3 months ago