Help understanding Torque Solution

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In summary, the conversation discusses determining the forces in each member of a truss and determining if the members are in tension or compression. The use of trigonometry (soh-cah-toa) is discussed to determine the x and y components of the forces. The direction of the forces is also discussed in relation to force balancing. The conversation ends with a question about the direction of one of the forces.
  • #1
NoobeAtPhysics
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Homework Statement



Determine the force in each member of the truss, and state if the members are in tension or compression.

2h58n6p.png


Homework Equations



T=rf

The Attempt at a Solution



I am trying to understand why they used sin for the x component of Joint D and cos for the y component of Joint D. I don't really understand when to use cos and when to use sin.
 
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  • #2
The mnemonic I was taught was sOH-cAH-tOa

sin(x) = opposite/hypotenuse
cos(x) = adjacent/hypotenuse
tan(x) = opposite/adjacent

(of course, you could simply derive the tOA from the sOHcAH)

Anyway, never believe in a universal "x" or "y" component when it comes to geometry. always base it off the angle in question and what's opposite or adjacent to it (or the hypotenuse).
 
  • #3
Thanks for the response

Hmm, I understand soh-cah-toa but I'm looking at Joint D and I just don't really understand.

If I use soh-cah-toa on joint D, then does sin(26.57) = 1/FDc?
 
  • #4
since the angle is from the top corner (I assume), then the opposite side is the x component and the hypotenuse is of course, FDC and that's what the book has.
 
  • #5
Ok, I understand that the opposite of the triangle would be Fec (x component) and hypotenuse if Fdc but we use sin because Fdc is the hypotenuse? I'm a bit confused.
 
  • #6
Yes Fec is equal to 0, so I don't really get how that decides when to use sin and cos
 
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  • #7
there is no Fec... I think you're confusing real space with vector space. We're looking at the components on the top of the scaffolding only. The length of Fec has nothing to do with it.
 
  • #8
I think I understand now! ( i think)

on a quick side note, would you know why Fde is pointing downward in the free body diagram for Joint D? Also is there a reason why Fdc is pointing up?

ad1bp4.png



I thought all joints were assumed to be in tension unless they are negative
 
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  • #9
the forces must sum to zero since it's not moving. Since the 600 are pushing from the left, the Fdc must counter it with a force from the right... and the arrow can't point down and right, so it must point left and right (given it's orientation). The only way to balance the resulting up force is by having Fde pull down.
 
  • #10
I don't understand you here:

'and the arrow can't point down and right, so it must point left and right (given it's orientation). The only way to balance the resulting up force is by having Fde pull down.'

I know that the Fdc should counteract the 600N but what does that have to do with Fde?
 
  • #11
At that joint, Fec is the only force up and Fde is the only force down. That's all they have to do with each other (through force balancing).

But it's all set in motion by the fact that 600 N are coming from the right, a horizontal force which must be countered by the horizontal component of Fdc, whose vertical component, then, must be countered by Fde. All in the name of force balance.
 
  • #12
If Fec is negative then

Fde should theoretically be positive to counteract the y force?edit:
Oh right!
 
  • #13
wait still, if Fde must be positive to counteract the y force, why is it pointing downwards?
 
  • #14
who said Fde must be up?
 

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