Resolve a force into three parts (3D)

[John Dough]

Hey!

I have been trying to solve this problem without success. Any help will be appreciated

Homework Statement

Resolve the 170 N force into three parts: one of which is parallel to OQ, another parallel to OP, and the third parallel to the y axis. (See the graph)
Are these the components of F in these directions?

Pay attention to the last question. the answer is "No". I have already found the components of the force in Fx, Fy and Fz directions. So the "parts" to be found are different from the components.

Homework Equations

Trigonometric equations
What I used to find the Fx, Fy and Fz components:
Fz = F * sinγ
Fy = F * cosγ * sinβ
Fx = F * cosγ * cosβ

The Attempt at a Solution

I found the Fx, Fy and Fz components using the equations above: they are 80, 120 and 90, respectively.

I also calculated the uni vector of F and of OQ (see attachment). I have also calculated the sides of several right triangles.
Now my problem is that I don't have an idea how to connect all the dots and complete the solution.

Thank you in advance!

 

[voko]

Since you have the x,y,z components of the force, and the unit vector of OQ (and presumably OP), projecting the force onto OQ and OP should be fairly straightforward.

 

[John Dough]

I also thought it will be straightforward, but I seem to be doing something wrong.

I have the answers and:

• the part parallel to the y-axis is 42.2 j N. I have absolutely no idea how to reach this answer. As mentioned about, I found Fy = 120
• the part parallel to OQ is 97.5 * unit vector of OQ, i.e. 97.5 * ( 3/13 i - 4/13 j + 12/13 k). I tried projecting the force F on OQ using the dot product, but I don't get the same answer.
• The same goes for the projection of F on OP. The answer is 122 * (8/17 i + 15/17 j) and I get a different number.

Any idea what I am doing wrong?

 

[voko]

Explain how you computed the relevant sines and cosines.

 

[John Dough]

It is visible from the IMG 0001.

Sin γ = 9 / 17 (right triangle with sides 14,422; 9 and 17. It is visible which side is 9, the hyp. is 17)
That's what I used to find Fz

Cos γ = 14.422 / 17 (same right triangle; 14.422 calculated from the rectangle in the XY-plane)

Sin β = 12 / 14.422 (14.422 is now the hypotenuse in a right triangle with sides 8 and 12)

Cos β = 8 / 14.422 (Adjacent / Hypotenuse)

 

[voko]

So the force is 170 * (8i + 12j + 9k) / 17 = 80i + 120j + 90k. This is correct. Your trigonometry is also valid.

I think what you are really required to find here is the decomposition of the force in the non-orthogonal basis given by the specified vectors.

 

[Chestermiller]

I also thought it will be straightforward, but I seem to be doing something wrong.

I have the answers and:

• the part parallel to the y-axis is 42.2 j N. I have absolutely no idea how to reach this answer. As mentioned about, I found Fy = 120
• the part parallel to OQ is 97.5 * unit vector of OQ, i.e. 97.5 * ( 3/13 i - 4/13 j + 12/13 k). I tried projecting the force F on OQ using the dot product, but I don't get the same answer.
• The same goes for the projection of F on OP. The answer is 122 * (8/17 i + 15/17 j) and I get a different number.

Any idea what I am doing wrong?

The problem is that the unit vectors in the OP, OQ, and y directions are not mutually orthogonal. So OP and OQ have components in the y direction. It isn't clear, but apparently, they are asking you to express F in terms of the non-orthogonal unit vectors in the OP, OQ, and y directions. That is,
\vec{F}=F_P\vec{i_p}+F_Q\vec{i_Q}+F_y\vec{i_y}
If you express the unit vectors for this equation in terms of the x, y, and z unit vectors, and then dot the resulting equation with each of the x, y, and z unit vectors, you end up with a set of three linear algebraic equations in three unknowns for F_P, F_Q, and F_y (this F_y is not the same as the component when resolving F into components in the x, y, and z directions). You then solve for these three unknowns.