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 \(3:2\sqrt { 2 }\)

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## Comments

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TopNewestLet us assume that the

resultant forceis 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. Thevertical componentof \(F_2\) 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.thehorizontal componentof \(F_2 = F_2/3\) . By simple trigonometry we get that \(\sin \theta = 1/3\) where \(\theta\) is the angle between theverticaland \(F_2\) . 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.Log in to reply

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.

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Can you make a effort to show the process?

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Dude its just pythagoras theorem.

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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....

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