Sum of the Forces at a point

  • #1
249
1

Homework Statement



If the weight of the boom is negligible compared with the applied 45-kN load, determine the cable tensions T1 and T2 and the force acting at the ball joint at A.


Homework Equations





The Attempt at a Solution



I have successfully calculated the tensions in T1 and T2 to be 71.1512 kN and 60.749kN respectively.

But I've been having a lot of difficulty calculating the forces at point A. My attempt is as follows:

ƩFx = 0

-T2 + T1cos(θx) + Ax

θx = arccos(9/15.81) where 15.81 is the length of the cable T1

Ax = 60.749 - 71.1512cos(55.304)
Ax = 20.2481

ƩFy = 0

θy = arctan (15/9)
θy = 59.036

0 = t1cos(59.036) - 45 + Ay
Ay = 8.393

ƩFz = 0
θz = arctan(10.2956/12) where 10.2956 is the length from A to C.
θz = 40.62857

0 = Az - 71.1512cos(40.62857)
Az = 53.999

A = [itex]\sqrt{}20.2481^2 + 8.393^2 + 53.999^2[/itex]
A = 58.278

It says my answer is wrong and I don't see where I'm going wrong. Help would be greatly appreciated.
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
Hi Northbysouth! :smile:

(try using the X2 button just above the Reply box :wink:)
ƩFx = 0

-T2 + T1cos(θx) + Ax

θx = arccos(9/15.81) where 15.81 is the length of the cable T1

Ax = 60.749 - 71.1512cos(55.304)
Ax = 20.2481

yes :smile:, but there was no need to find the actual angle

all you needed to know in -T2 + T1cos(θx) + Ax was cosx) …

which you know is 9/15.81 ! :wink:

i don't understand why you then used tan for the y and z directions, instead of the cos you used for the x direction :confused:

try it the same way :smile:

(if it helps, you can visualise the tension T1 as replaced by forces of relative magnitudes 9:5:-12 along the x y and z axes)
 
  • #3
249
1
I'm not sure I completely understand what you're saying.

When I was calculating the angles I imagined that the origin was at point B. From there I then used trigonometry to determine what angle the force T1 made with each axis.

So, when I was determining θy I visualized a right angle triangle that went from B to C and then directly back to the y axis at B.

Are you saying that I should use arccos instead of arctan when I'm calculating the angles at θy and θz?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
251
So, when I was determining θy I visualized a right angle triangle that went from B to C and then directly back to the y axis at B.

not following you :confused:

let's see what you did …
θy = arctan (15/9)

where does the 15 come from?

(i can see a 15, but it's a hyptoneuse, and it's the wrong angle)

the unit vector along T1 is (9,5,-12)/15.81 …

the cosines are the individual coordinates :smile:
 
  • #5
249
1
I understand what you mean when you said that I didn't need to calculate the angles.

Using the unit vector of T1 = <9, 5, -12>/[itex]\sqrt{}250[/itex]

I multiplied the components i, j and k by the force to get the magnitude of the force in the x, y and z direction. From there it was simple to calculate Ax, Ay and Az

The answer I got for A is 61.9053 kN, which is correct.

Thank you for your help.
 

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