physicsvectors
Forces A and B has a resultant force C with magnitude of 200N.
The magnitudes of A and B have the relation of 2||A||=3||B||.
$\theta$ is the angle between A and C, and the angle between B and C is 2$\theta$.
Determine the value of $\cos\theta$ and $\cos2\theta$.
Then find ||A|| and ||B||.
General tips to get me started on this is appreciated.
1) Draw a diagram with coordinate axes:
2) Determine the components of the two forces:
$${F_A}_x = F_A = 150 lbs \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space {F_A}_y = 0$$ $$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space{F_B}_x = -F_B\cos70° = -103 lbs \space\space\space\space\space\space\space\space\space\space{F_B}_y = F_B\sin70° = 282 lbs$$ 3) Find the components of the resultant force $$R_x = {F_A}_x + {F_B}_x = 47 lbs$$ $$R_y = {F_A}_y + {F_B}_y = 282lbs$$ 4) Apply Pythagoras' theorem to get the magnitude of the resultant: $$R = \sqrt {R_x^2 + R_y^2} = 286lbs$$
If you want to know more, here's a good article which explains each step in more detail: How to find the resultant force acting on an object
In general, if $\vec{x} = a\vec{i}+b\vec{j}$, then the magnitude of $\vec{x}$, denoted $\lVert \vec{x}\rVert,$ is given by $\lVert \vec{x}\rVert = \sqrt{a^2+b^2}$. Does this help?
Best Answer
Hint: Given the angles, the angle between $A$ and $B$ will be $3\theta$ (As $A,B,C$ must lie in a plane). We have $$\tag12|A|=3|B|$$ $$\tag2|C|=\sqrt{|A|^2+|B|^2+2|A||B|\cos3\theta}$$
If the angle between $A$ and $C$ is $\alpha$ and that between $A$ and $B$ is $\theta$, then $$\tag3\tan\alpha=\frac{|B|\sin\theta}{|A|+|B|\cos\theta}$$ Here, replace $\alpha$ with $\theta$ and $\theta$ with $3\theta$. $$\tag4|C|=200$$
Use $(1)$, $(3)$ and $(4)$ to substitute in $(2)$