# Vectors Algebra

A boat is being pulled by two men.The stronger men exerts a force which is $(3/2)^{1/2}$ times that of the weaker man, and in the direction at angle $45^{\circ}$ north of east (i.e. $45^{\circ}$ counterclockwise from the positive $x$-axis). Which direction must the weaker man pull to ensure the boat travels due east(positive $x$ direction)?

The answer given is $60^{\circ}$.

How to solve this type of question. Could someone give me some ideas to solve it?

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Hint: Set the $y$-components of the vectors equal to each other and solve for the unknown angle. –  David Mitra Sep 17 '12 at 16:19
Thanks david mitra –  Garett Sep 17 '12 at 17:31

If the weaker man's force is $F$ at a direction $\theta$ south of east, then the resultant force on the boat is $(\sqrt{\frac{3}{2}}F \cos 45^\circ+F \cos \theta, \sqrt{\frac{3}{2}}F \sin 45^\circ-F \sin \theta) = F (\sqrt{\frac{3}{2}}\cos 45^\circ+\cos \theta, \sqrt{\frac{3}{2}} \sin 45^\circ- \sin \theta)$.

In order for the boat to go in the $x$-direction, you need the $y$ resultant to be zero, ie, $\sqrt{\frac{3}{2}}\sin 45^\circ =\sin \theta$, which gives $\sin \theta = \frac{\sqrt{3}}{2}$, or $\theta = 60^\circ$. The resultant force on the boat is $F(\frac{\sqrt{3}+1}{2},0)$.

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YES!!! I got it. Thank you very much bro –  Garett Sep 17 '12 at 17:30

Let the magnitude of the force exerted by the weaker person be $w$. Then the magnitude of the force exerted by the stronger one is $w\sqrt{\frac{3}{2}}$.

Suppose that the weaker person will be exerting force in a direction $\theta$ degrees South of East. Make an informal sketch.

In order to make sure that the resultant force goes due East, the northward component of the force exerted by the stronger person must be cancelled by the southward component of the force exerted by the weakling. It follows that $$w\sqrt{\frac{3}{2}}\sin(45^\circ)=w\sin\theta.$$

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ThANKS for your help Andre –  Garett Sep 17 '12 at 17:30