Pulling a rope on a wall, finding gamma

[Bluestribute]

Homework Statement

Determine coordinate direction angle γ of the resultant force acting at B.

Homework Equations

Cos$\gamma$ = F$\underline{}z$/F

The Attempt at a Solution

The first part was to find magnitude of F$\underline{}b$. The next part was finding the other coordinate directions. I found all but gamma. I don't know how gamma is wrong but the rest are correct. My correct numbers (rounded from the website):

F$\underline{}B$ = 78.9 lb
$\alpha$ = 35.4°
$\beta$ = 68.8°

Why won't the same method work for finding $\gamma$ (the method being the equation up there)? I keep getting 63.1°.

Components of F$\underline{}B$: <69.30 , 28.55 , 35.75>

 

[STEMucator]

Your $\alpha$ and $\beta$ angles make sense. That is not the correct magnitude of the force at $B$.

I get $F_B = 110 lb$. Could you show some of your working so we can see where you went wrong?

 

[Bluestribute]

That's weird because, except for rounding, it said my answers were correct (btw, gamma was 117° and I don't know how . . . ).

uF$_{}c$ = (6/14)i + (12/14)j + (4/14)k
F$_{}c$ * 50 (the force) = 21.45i + 42.85j + 14.30k

F$_{}a$ I did the same, but got this overall vector:
42.85i - 14.30j + 21.45k It had a negative j because it's left of the point "B" in the picture, on the y axis.

Then I added those to get the resultant F$_{}B$ vector, 64.30i + 28.55j +35.75k. Finding the magnitude (root of the squares) gets me 78.92, which is correct on the website . . .

 

[STEMucator]

Hmm i seem to be getting :

$F_B = \{90 \hat i + 40 \hat j - 50 \hat k \} N$

How are you getting $\vec F_A$ and $\vec F_C$? You should be finding position vectors $\vec r_A$ and $\vec r_C$, which you then normalize to obtain unit vectors $\vec u_A$ and $\vec u_C$.

You can then express the forces $\vec F_A$ and $\vec F_C$ in terms of the unit vectors.

Also, you should list the question as stated in the book with all the relevant variables. I'm sure magnitudes for $F_A$ and $F_C$ have already been provided.

 

[Bluestribute]

Yes they are 50 lb. I thought I was posting a continuation of a problem. Forgot it was different!

I'm using the coordinates in the picture. You can see in my Fc equation what my position and magnitude were based solely off the picture.

 

[STEMucator]

Well If you have them as $50 lb$ each:

$\vec F_A = F_A \vec u_A = \{ \frac{300}{7} \hat i + \frac{100}{7} \hat j - \frac{150}{7} \hat k \} N$

$\vec F_C = F_C \vec u_C = \{ \frac{150}{7} \hat i + \frac{300}{7} \hat j - \frac{100}{7} \hat k \} N$

I get the same vectors as you. So I'm assuming your $F_B$ is correct now that we have the magnitudes of $F_A$ and $F_C$.

All you would need to do is use the coordinate angle relationship now to find the three angles.

 

[Bluestribute]

Right. And I did to get alpha and beta. Gamma gave me 63 I believe. Gamma was 117 and I don't know how they calculated that since I used the same numbers and same procedure to get alpha and beta (which were correct).

Is there a special way to know when the angle you get needs to be subtracted from 180 (I got 63, the website got 117, 180 - 63 = 117)?

 

[STEMucator]

Right. And I did to get alpha and beta. Gamma gave me 63 I believe. Gamma was 117 and I don't know how they calculated that since I used the same numbers and same procedure to get alpha and beta (which were correct).

Is there a special way to know when the angle you get needs to be subtracted from 180 (I got 63, the website got 117, 180 - 63 = 117)?

Ask yourself which way $\vec F_{B_z}$ points. You should find this changes your answer to $117 °$.

 

[Bluestribute]

I could ask myself that and still not know . . . I can't imagine gamma being obtuse until maybe I see an example probably . . .

 

[STEMucator]

I could ask myself that and still not know . . . I can't imagine gamma being obtuse until maybe I see an example probably . . .

What do you mean? Just observing the drawing indicates a net downward force in the $z$ direction at $B$.

 

[Bluestribute]

So if it's going down the Z axis, it's going to be an obtuse angle for gamma?

 

[STEMucator]

I wind up computing $\cos^{-1}(\frac{- 35.71}{78.9})$. The $z$ component of $B$ points downwards.

 

[Bluestribute]

Alright, I think I see where I went wrong. This has been way more helpful than class.