Finding Angle θ between 300-lb Force and Line OC

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Discussion Overview

The discussion revolves around determining the angle θ between a 300-lb force and the line OC in a three-dimensional coordinate system. Participants explore vector calculations and the relationships between different points in the problem, addressing both the mathematical approach and potential errors in their calculations.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant initially calculates the angle between vector AB and OA, mistakenly thinking it was the required angle.
  • Another participant points out that the angle should be calculated with respect to OC instead.
  • Several participants share their recalculations for the angle with OC, resulting in different values, indicating potential discrepancies in their methods or calculations.
  • One participant suggests that the calculations for OC might be incorrect and prompts a review of the vector definitions.
  • After corrections, a participant arrives at a new angle of 60.42°, claiming it is correct, while others express ongoing confusion about earlier results.

Areas of Agreement / Disagreement

There is no consensus on the correct angle until the final post, where one participant claims to have arrived at the correct answer. However, earlier posts reflect multiple competing calculations and interpretations of the vectors involved.

Contextual Notes

Participants express uncertainty regarding the correct definitions of vectors and the calculations involved, indicating that attention to detail is crucial in solving the problem. Some calculations appear to depend on earlier assumptions that may not have been clearly defined.

Northbysouth
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Homework Statement



Determine the angle θ between the 300-lb force and the line OC.

I have attached an image of the question


Homework Equations



P°Q = PQcos(θ)



The Attempt at a Solution



I tried finding the vectors AB and OA as well as their magnitudes first
O is the origin
A = (9,0,0)
B = (0, 16, 6)
C =(9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-92) + 162 + 62)
|AB| = 19.31

OA = <9-0, 0-0, 0-0>
OA = <9, 0, 0>

|OA| = 9

AB°OA = |AB||OA| cos(θ)

<-9, 16, 6>°<9, 0, 0> = (19.31)(9)cos(θ)

arccos(-81/173.79)

θ = 117.78

It says this answer is not correct and I realize that I haven't included the 300-lb force, but how would I take that force into account?
 

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Looks like you've found the angle of AB with line OA ... but it asks for the angle with OC.
 
...Excellent observation...it's been a long day.

So I ran the calculations again, this time with OC. and I got an angle of 34.046 which is still wrong. :/
 
I ran the calculations, for the angle with OC using the given info and got a larger angle than yours, double check your calculations.
 
So I checked my calculations and still got 34.056°. My calculations are as follows:

O is the origin
A = (9,0,0)
B = (0, 16, 6)
C =(9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-92) + 162 + 62)
|AB| = 19.31

OC = <9-9, 16-0, 0-0>
OC = <0, 16, 0>

|OC| = 16

AB°OC = |AB||OC| cos(θ)

<-9, 16, 6>°<0, 16, 0> = (19.31)(16)cos(θ)

θ = arccos(256/308.96)

θ = 34.046
 
Shouldn't OC = <9,16,0> ?
 
Yes, you are correct. I think I've been at this too long :/.

My calculations are as follows

A = (9, 0, 00
B = (0, 16, 6)
C = (9, 16, 0)

AB = <0-9, 16-0, 6-0>
AB = <-9, 16, 6>

|AB| = sqrt((-9)^2 + 16^2 + 6^2)
|AB| = 19.31 kN

OC = <9-0, 16-0, 0-0>
OC = < 9, 16, 0>

|OC| = sqrt(9^2 + 16^2)
|OC| = 18.358 kN

AB°OC = |AB||OC|cos(θ)

<-9, 16, 6>°<9, 16, 0> = (19.31)(18.358) cos(θ)

θ = arccos(175/354.47)
θ = 60.42°

This answer is correct. Thanks for all your help.
 
There's been a bunch of people struggling through these and it is usually just down to attention to detail. It gets everybody like that - you'll get used to it.
 

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