Four forces of magnitudes P, 2P, 3P and 4P act along four sides of a square, calculate resultant force.

Question

Four forces of magnitudes P, 2P, 3P and 4P act along the four sides of a square ABCD in cyclic order. Use the vector method to find the resultant force.

I tried solving this question in two ways and I got two different answers which differ only by a very small magnitude of approximately 0.1.

First I solved it in the following way: Let the vectors P, 2P, 3P and 4P be along the sides AB, BC, CD and DA respectively. Now, the resultant of P and 2P will be $\sqrt{P^2 + 4P^2}$ = $\sqrt{5}$P. Resultant of 3P and 4P will be $\sqrt{9P^2 + 16P^2}$ = 5P. Because the resultant of 3P and 4P is greater than that of P and 2P so I subtracted and got the final result as 5P - $\sqrt{5}$P which equals to 2.763P.

Then I solved it in the following way: First I re-arranged them like this and then I will get two vectors with the same magnitude 2P. Their resultant will be 2$\sqrt{2}$P which equals to 2.828P.

So I am basically struggling to understand that why this small difference in the answers is arising when changing between the two methods. Please help me to know where I am making a mistake, thank you.

Answer 1

So your first answer is wrong, the second is correct. The fact that they’re only off by a little bit is mostly coincidence.

Here’s your mistake: when you calculated the resultant of the vectors of magnitudes P and 2P you got the magnitude of the resultant, but not the direction. Then when you did the same with the vectors of magnitudes 3P and 4P, you also only got the magnitude of the resultant.

Now these vectors do not point in the same direction. So you cannot simply add their magnitudes to get the resulting vector. That was your error.

Your second attempt was perfect! In general, when you have a problem with vectors and you’re given them in “coordinate form” as in an (x,y) pair, keep them in that form for all your calculations. Putting them in “direction, magnitude” form makes even simple vector addition very complicated and introduces unnecessary square roots and trigonometry.

And of course, if the question requires it, you can always put your final answer as a direction and a magnitude.

Answer 2

To obtain the result we can proceed at least in two ways.

Geometrically we can use the tip to tail method which according to the following sketch

gives $2\sqrt 2 P$ as a solution (vector $\vec{AE}$).

As an alternative, algebraically by coordinates we have

• $\vec {AB}=(0,-P)$
• $\vec {BC}=(-2P,0)$
• $\vec {CD}=(0,3P)$
• $\vec {DE}=(4P,0)$

therefore

$$\vec {AE}=\vec {AB}+\vec {BC}+\vec {CD}+\vec {DE}=(2P,2P) \implies \left|\vec {AE}\right|=2\sqrt 2 P$$

Answer 3

This is easier if you add vectors on opposite sides of the square first. $P$ is being opposed by $3P$ and $2P$ is opposed by $4P$.

Assume $P$ points to the right and $2P$ points up. Then we get $$P-3P=-2P$$ in the $x$ direction and $$2P-4P=-2P$$ in the $y$ direction.

Therefore, the resultant force is $$-2P\hat i+-2P\hat j$$