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What I don't understand in this picture is why they are taking $$5\cos130^{\circ}$$ I usually do these problems by drawing the components along the axes in my case I would draw the y component along the y-axis and the x component would be going across on the top from the y axis to the resultant vector.

If I draw my right triangle like that then would the x component would be $$5\sin130^{\circ}$$ I do see though that this way would give me an incorrect x component but I don't know why they are using cosine as if I drew it on paper the angle that is between the hypotenuse and base would not $$=130^{\circ}$$

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Another way of understanding the cosine is through the definition of the scalar product $$ \vec{a}\cdot\vec{b} := |\vec{a}||\vec{b}|\cos\angle(\vec{a},\vec{b}) $$ If you want to find the x component $B_x$ of your vector $\vec{B}$, you just have to multiply with the unit vector in x direction, this is called $\hat{i}$ in your drawing. This way you get $$ B_x = \vec{B}\cdot\hat{i}=|\vec{B}|\underbrace{|\hat{i}|}_{=1}\cos(130°)=5\cos(130°) $$ Similarly you get for $B_y$ $$ B_y = \vec{B}\cdot\hat{j}=|\vec{B}||\hat{j}|\cos(40°)=5\sin(130°) $$

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  • $\begingroup$ I understand this concept and it helped $\endgroup$
    – Jude
    Mar 6 '17 at 1:56
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In order to find the x-component of a vector, you take the cosine of the angle and multiply it by the magnitude of the vector. In this case, taking $5\cos{130^\circ}$ is correct to find $\vec{B}_x$.

Your intuition is somewhat correct, but in order to do what you've done by drawing the x-component from the y-axis to the end of $\vec{B}$ you need to use the correct angle. Since $90^\circ$ of the $130^\circ$ is already "used" in the first quadrant, you need to subtract 90 degrees from 130 to get the correct angle to use sine with. $5\sin{180^\circ}$ results in the same answer.

You don't want to do the above on a test or paper because it is unclear where you got your numbers from unless you explicitly state it. Stick with cosine for the horizontal component of the vector because that is the most common standard.

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  • $\begingroup$ for some reason $5\sin40^{\circ}$ is not giving the proper x component value $\endgroup$
    – Jude
    Mar 6 '17 at 0:20
  • $\begingroup$ What value is it supposed to be? It should be the same as $5\cos{130^\circ}$ $\endgroup$
    – Artillect
    Mar 6 '17 at 0:27
  • $\begingroup$ Wait, I'm wrong. This is basically why you should never use sine instead of cosine to find the horizontal component of the vector. Just stick with using cosine. It should be $5\sin{220^\circ}$ $\endgroup$
    – Artillect
    Mar 6 '17 at 0:37
  • $\begingroup$ So no matter how my right triangle could end up cosine should always be used for horizontal? It just seems so odd when I picture the triangle since in the books case the $130^{\circ}$ would be on the outside of the triangle I just don't see how you can take the adjacent/hypotenuse $\endgroup$
    – Jude
    Mar 6 '17 at 1:59
  • $\begingroup$ Don't imagine the cosine as the function you use to solve a right triangle in contexts like this one. When you're finding the horizontal component of the vector, you're finding the dot product of your vector and $\hat{\textbf{i}}$, which is given by $\vec{\textbf{A}} \textit{\hat{i}}} = \abs{\vec{\textbf{A}}\cos{\theta}$. $\endgroup$
    – Artillect
    Mar 6 '17 at 3:53

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