How can the magnitude of resultant force of $(0,3)$ and $(3,1)$ be doubled by changing $(0,3)$ but keeping $(3,1)$ unchanged.

Question

How can the magnitude of resultant force of $(0,3)$ and $(3,1)$ be doubled by changing $(0,3)$ but keeping $(3,1)$ unchanged.

since magnitude of resultant force (3,4) is $\sqrt{25}$. Double of magnitude will be sq root 50... Now what should be done i am confused here...

Answer 1

Your question isn't very clear, but if you are trying to double the magnitude of $(0,3)+(3,1) = (3,4)$ by replacing only the first vector, then you need to solve the following equation : $|(x,y)+(3,1)|= |(3+x,1+y)|=\sqrt{(3+x)^2 + (1+y)^2} = 2\sqrt{25}$. This an equation with two variable, you will have more then one solution. Pick one and you're set.

Good luck.

Answer 2

You have

• $\vec F_1 = \begin{pmatrix} 0 \\ 3 \end{pmatrix}, \vec F_2 = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \Rightarrow \vec F_r = \vec F_1 + \vec F_2 = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
• $|\vec F_r| = \sqrt{3^2 + 4^2} = 5$

If I understand you correctly , you want to find a positive number $t$ such that for vector $$\boxed{\vec F_t = t \vec F_1 + \vec F_2} = \begin{pmatrix} 3 \\ 3t+1 \end{pmatrix}$$ holds

$$|\vec F_t| = |t \vec F_1 + \vec F_2| = 2 |\vec F_r| = 10$$

So, set up the equation $$3^2 + (3t+ 1)^2 = 10^2 = 100 \stackrel{t > 0}{\Rightarrow} \boxed{t = \frac{1}{3}\left( \sqrt{91} - 1\right)}$$