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Vector $\vec{V} \!_1$ is 6.6 units long and points along the negative x axis. Vector $\vec{V} \!_2$ is 8.5 units long and points at $+55^\circ$ to the positive x axis.

  1. What are the x and y components of each vector?
  2. Determine the sum $\vec{V} \!_1 + \vec{V} \!_2$ (magnitude and angle).

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 
  1. $V_{1_x} = -6.6, V_{1_y} = 0, V_{2_x} = 4.9, V_{2_y} = 7.0$
  2. $\vec{V}=7.2 \textrm{ units } 76^\circ \textrm{ above neg. x-axis}$

Giancoli 7th Edition, Chapter 3, Problem 6


Chapter 3, Problem 6 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. Vector 1 is easy to resolve into components because it's entirely in the x-direction. So its x-component is negative 6.6 if we take to the right being positive and up being positive here and the y-component of vector 1 is zero. Vector 2, its x-component is gonna be the length of this hypotenuse 8.5 and I'm thinking about this triangle here 8.5 divided by or sorry times cos 45 because cos 45 or 55 I should say cos 55 is the adjacent which is v 2 x divided by v 2 and so we are mulitplying both sides by v 2 here and then solving for v 2 x so v 2 x is v 2 times cos 55. So that's 8.5 times cos 55 which is about 4.9 units and that's to the right so it's positive. v 2 y is same algebra and idea but with sin now and then that gives about 7.0 units upwards. And then in part (b) find the magnitude to the resultant; well, we have to find the total x-component of the resultant and that's the x-component of the first vector plus the x-component of the second vector so that's negative 6.6 plus 4.9 which is negative 1.7 units. And then v y is the sum of the y-components and that is just that of vector 2 which is 6.9628 and then the length of the resultant vector is the square root of the sum of the squares and that gives us 7.167 units and then we'll take the inverse tangent of the y-component over the x-component of the resultant to— and so we use the resultant by the way and put that in the picture— and the resultant has this x-component here, that's v R x and then it has this y-component here, v R y component and v R x is negative 1.7 and v R y is 6.9628 and then the angle Θ which was gonna be tucked in here the inverse tangent of this opposite divided by the adjacent will give us the angle Θ. So that Θ, according to our picture, is gonna be above the negative x-axis. And so here we have it; 7.2 units, 76 degrees above the negative x-axis is our resultant.