Determine the components of P in the i, j, and k directions.

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In summary, P is a force that is projected along the x axis. The x component of P is zero if alpha is 90 degrees or 270 degrees.
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anonymous812
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Homework Statement


As shown, a pole is subjected to three forces: F, P, and T. The force F expressed in Cartesian vector form is F=[5 i+7 j+6 k] N. Force P has magnitude 7.48 N and acts in the direction given by these direction angles: α=143∘, β=57.7∘, and γ=74.5∘ to the x, y, and z axes, respectively. Force T lies within the yz plane, its direction is given by the 3-4-5 triangle shown, and its magnitude is 14 N.
Determine the components of P in the i, j, and k directions.

Homework Equations


A_x=Acos(α)
cos^2(α)+cos^2(β)+cos^2(λ)=1


The Attempt at a Solution


Honestly, I'm very confused as to how to approach this problem. I was under the impression that the x component was 0, y component was 8.40 and the z component was 11.2. I solved these by making a 2D plane from the P vector into a 3-4-5 right triangle. 3 on y, 4 on z, and 5 as the hyp.
I would really appreciate a better explanation of how to approach this question..I think I'm making it more difficult than it is.
Thank you!
 

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  • #2
Why arer F and T given if P is the only force to be found? It's not as if P is somehow related to T and/or F.
 
  • #3
There are multiple parts to the question.. I'm only asking about the P vector.
 
  • #4
anonymous812 said:
1. . I was under the impression that the x component was 0, y component was 8.40 and the z component was 11.2. I solved these by making a 2D plane from the P vector into a 3-4-5 right triangle. 3 on y, 4 on z, and 5 as the hyp.
I would really appreciate a better explanation of how to approach this question..I think I'm making it more difficult than it is.
Thank you!


Taking just the x component of P, the only way that component could be zero is if alpha = 90 (or 270) deg.

Look at P: what is the projection of P along the x axis?
Hint: cos(α) = P*i/|P|

(vectors in bold, * signifies dot product)
 
  • #5


To determine the components of P in the i, j, and k directions, we can use the given magnitude and direction angles to calculate the x, y, and z components of P.

First, we can use the formula A_x = A*cos(α) to find the x component of P. Plugging in the values, we get A_x = 7.48*cos(143°) = -2.69 N.

Next, we can use the formula A_y = A*cos(β) to find the y component of P. Plugging in the values, we get A_y = 7.48*cos(57.7°) = 4.77 N.

Lastly, we can use the formula A_z = A*cos(γ) to find the z component of P. Plugging in the values, we get A_z = 7.48*cos(74.5°) = 2.63 N.

Therefore, the components of P in the i, j, and k directions are -2.69 N, 4.77 N, and 2.63 N, respectively.

It is important to note that while your approach of using a 3-4-5 right triangle to solve for the components may work, it is not a generalizable method and may not be accurate for other vectors with different magnitudes and direction angles. Using the formulas for calculating the components is a more reliable and efficient method.
 

FAQ: Determine the components of P in the i, j, and k directions.

1. What is the meaning of "components of P in the i, j, and k directions"?

When we talk about the components of P in the i, j, and k directions, we are referring to the vector components of a vector P in a three-dimensional coordinate system. The i, j, and k directions correspond to the x, y, and z axes respectively, and the components represent the magnitude of P in each of these directions.

2. How do you determine the components of a vector P in the i, j, and k directions?

To determine the components of a vector P in the i, j, and k directions, you can use the dot product or cross product of P with the unit vectors i, j, and k. The dot product gives you the scalar components of P in each direction, while the cross product gives you the vector components.

3. What is the significance of determining the components of a vector in multiple directions?

Determining the components of a vector in multiple directions allows us to break down a complex vector into simpler components, making it easier to analyze and understand. It also helps in calculating the magnitude and direction of the vector more accurately.

4. Can the components of a vector in the i, j, and k directions be negative?

Yes, the components of a vector in the i, j, and k directions can be negative. This indicates that the vector is pointing in the opposite direction of that particular axis. For example, a vector with a negative i component would be pointing in the negative x direction.

5. What is the difference between the i, j, and k components of a vector and the magnitude and direction of the vector?

The i, j, and k components of a vector represent the vector's projection onto the x, y, and z axes respectively. On the other hand, the magnitude and direction of a vector describe the length and orientation of the vector in space. While the components are numerical values, the magnitude and direction are expressed as a scalar and an angle, respectively.

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