# Resolve a force into three parts (3D)

1. Jul 14, 2013

### John Dough

Hey!

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

1. The problem statement, all variables and given/known data
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.

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

3. 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.

#### Attached Files:

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• ###### IMG000.pdf
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2. Jul 14, 2013

### 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.

3. Jul 15, 2013

### John Dough

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

• 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?

4. Jul 15, 2013

### voko

Explain how you computed the relevant sines and cosines.

5. Jul 15, 2013

### 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)

6. Jul 15, 2013

### 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.

7. Jul 15, 2013

### Staff: Mentor

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 FP, FQ, and Fy (this Fy 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.