Magnitude and direction of Vectors using head to tail rule

In summary, the conversation discusses a homework problem involving the calculation of resultant forces using the x and y components. The initial attempt at solving the problem is incorrect due to a mistake in the direction of the x-component. The correct solution is then provided and confirmed, with an additional discussion about the angle of the resultant force.
  • #1
bllnsr
26
0

Homework Statement


j8ei6u.png



Homework Equations



[itex]F = \sqrt{F_x^2 + F_y^2}[/itex]
[itex]tan\theta = F_y / F_x[/itex]

The Attempt at a Solution



Page 1 : http://i49.tinypic.com/vfw74k.jpg
Page 2 : http://i50.tinypic.com/2qspamr.jpg
Page 3 : http://i49.tinypic.com/24zgl6x.jpg
is it correct?
 
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  • #2
I think you've got the directions wrong. The x-component of forces are in opposite directions, so you need to account for that.

You've calculated the y-component of the resultant correctly, but even there you haven't specified the downward direction which could cause you to lose marks sometimes.

Since, your calculation for x-components is wrong, the magnitude and direction of the resultant is also wrong...:smile:
 
  • #3
the first two look okay the last page #3 you used vector length value so your tan angle is wrong.

A quick way to check your solution is to graph it and take measurements for the angle and length to see that your results agree.
 
  • #4
jedishrfu said:
the first two look okay the last page #3 you used vector length value so your tan angle is wrong.

Shouldn't Fx=-6000cos(60°)+2000cos(45°)?
 
  • #5
MrWarlock616 said:
Shouldn't Fx=-6000cos(60°)+2000cos(45°)?

Yes, I think you're right. I didn't notice that.
 
  • #6
so after doing Fx=-6000cos(60°)+2000cos(45°)
resultant force is 6797.915 correct?
 
  • #7
bllnsr said:
so after doing Fx=-6000cos(60°)+2000cos(45°)
resultant force is 6797.915 correct?

Yes. And what will be the angle made by that resultant with the positive x-axis??
 
  • #8
[itex]tan\theta = f_y/F_x[/itex]
76.51(i didn't use minus sign in f_y)
180-76.51= 103.48
right?
 
  • #9
bllnsr said:
[itex]tan\theta = f_y/F_x[/itex]
76.51(i didn't use minus sign in f_y)
180-76.51= 103.48
right?

Correct! :)
 
  • #10
Thanks for the help
 

Related to Magnitude and direction of Vectors using head to tail rule

1. What is a vector?

A vector is a quantity that has both magnitude (size or length) and direction. It is represented by an arrow, where the length of the arrow represents the magnitude and the direction of the arrow represents the direction.

2. How do you determine the magnitude of a vector?

The magnitude of a vector can be determined by using the Pythagorean theorem, which states that the magnitude is equal to the square root of the sum of the squares of the vector's components. For example, if the vector has components (x, y), the magnitude would be the square root of (x^2 + y^2).

3. What is the head-to-tail rule for adding vectors?

The head-to-tail rule states that the resultant vector of two or more vectors can be determined by placing the tail of one vector at the head of the other vector, and drawing a new vector from the tail of the first vector to the head of the last vector. The direction and magnitude of the resultant vector can be determined using trigonometry.

4. Can vectors be negative?

Yes, vectors can have negative components, which means they can point in the opposite direction of the positive axis. This is represented by a negative sign in front of the component value.

5. How do you subtract vectors using the head-to-tail rule?

To subtract vectors using the head-to-tail rule, you can simply reverse the direction of the vector that you are subtracting and then add the vectors using the head-to-tail rule. This will result in a new vector that is the difference between the two original vectors.

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