Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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 $F(\frac{\sqrt{3}+1}{2},0)$.

share|cite|improve this answer
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 $$w\sqrt{\frac{3}{2}}\sin(45^\circ)=w\sin\theta.$$

share|cite|improve this answer
ThANKS for your help Andre – Garett Sep 17 '12 at 17:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.