Need help with components of vectors

In summary, the problem involves finding the magnitude and angle of a vector C such that 2A + C - B results in a vector with a magnitude of 37 lbs pointing in the +x direction. Using the method of components, the calculated magnitude of C is 70.21 lbs and its direction is 333 degrees counterclockwise from the +x-axis. There may have been confusion with the angles of the vectors, but the solution is correct.
  • #1
godawgs
1
0

Homework Statement


1. Given the following force vectors: A is 27 lb at an angle of 27° clockwise from the +x-axis, and B is 44 lb at an angle of 45° clockwise from the +y-axis.

(a) Make a sketch and visually estimate the magnitude and angle of the vector C such that 2A+ C − B results in a vector with a magnitude of 37 lb pointing in the +x direction. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Repeat the calculation in Part (a) using the method of components and compare your result to the estimate in (a).
C=_______ lb
magnitude _______ lb
direction_______° counterclockwise from the +x-axis

Homework Equations


The Attempt at a Solution



Comp Ax 27cos(27) --> 24.1
Comp Ay 27sin(27) --> 12.3

Comp Ay 44cos(45) --> 31.1
Comp By 44sin(45) --> 31.1

Cx = Ax + Bx == 24.1 + 31.1 = 55.2
Cy = Ay + By == 12.3 + 31.1 = 43.4

C^2 = A^2 + B^2 == 55.2^2 + 43.4^2 == 4930.6
C = sqrt(4930.6) == 70.21 lbs

I attempted to add the components of the two vectors and then use the pythagorean theorem to find the magnitude of C but I feel like I'm missing a step. I think adding 3 different vectors is throwing me off... Can anyone help or walk me through the steps of the problem?
 
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  • #2
Hey! We're probably in the same class. I think you may have mistaken the degrees for your B vector. Since it is 56 degrees clockwise from the +y axis, it is actually 270 degrees + 56 degrees. Thus your components should be 44cos(315). Give that a try. Good luck!
 
  • #3
I'm not sure where fenderchik got 56[itex]^\circ[/tex] for vector B. Your problem states 45[itex]^\circ[/tex] from the positive y-axis (which lies at 90[itex]^\circ[/tex]).

Assuming your numbers are correct, you mis-judged the actual angle of vector A.
The positive x-axis lies at 0[itex]^\circ[/tex], so 27[itex]^\circ[/tex] clockwise from that is -27[itex]^\circ[/tex], or 333[itex]^\circ[/tex] (not +27[itex]^\circ[/tex]).

Otherwise, your work looks good.
 

FAQ: Need help with components of vectors

1. What are the components of a vector?

The components of a vector are the magnitude and direction. The magnitude is the size or length of the vector, while the direction is the angle at which the vector is pointing.

2. How do you find the components of a vector?

To find the components of a vector, you can use trigonometric functions such as sine, cosine, and tangent. The magnitude can be found using the Pythagorean theorem, while the direction can be determined by taking the inverse tangent of the y and x components of the vector.

3. What is the difference between a scalar and a vector?

A scalar is a quantity that has only magnitude, while a vector has both magnitude and direction. For example, temperature is a scalar quantity, while velocity is a vector quantity.

4. How are vectors used in physics?

Vectors are used in physics to describe quantities that have both magnitude and direction, such as force, velocity, and acceleration. They are also used in vector operations, such as addition, subtraction, and multiplication, to solve problems in mechanics and other areas of physics.

5. Can you give an example of a vector in everyday life?

Yes, a common example of a vector in everyday life is a person walking. The person's velocity can be represented as a vector with a magnitude (speed) and direction (forward or backward). Another example is a force applied to an object, which can also be represented as a vector with a magnitude and direction.

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