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
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Let us assume that the resultant force is in the horizontal direction and the force F1 in the vertical direction. The direction of F2 will be as shown in the figure. The vertical component of F2 should be equal to F1 since there is no resultant force in vertical direction. It is given that F2=Fr/3 i.e. the horizontal component of F2=F2/3 . By simple trigonometry we get that sinθ=1/3 where θ is the angle between the vertical and F2 . and therefore, cosθ=22/3 . Therefore, again by trigonometry, F2=F1×22/3 Therefore, we get the required ratio !! i'm not able to post a figure now, will post it later.
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....