Help understanding Torque Solution

1. Nov 20, 2013

NoobeAtPhysics

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

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

2. Relevant equations

T=rf

3. 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.

2. Nov 20, 2013

Pythagorean

The mnemonic I was taught was sOH-cAH-tOa

sin(x) = opposite/hypotenuse

(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. Nov 20, 2013

NoobeAtPhysics

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. Nov 20, 2013

Pythagorean

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. Nov 20, 2013

NoobeAtPhysics

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. Nov 20, 2013

NoobeAtPhysics

Yes Fec is equal to 0, so I don't really get how that decides when to use sin and cos

Last edited: Nov 20, 2013
7. Nov 20, 2013

Pythagorean

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. Nov 20, 2013

NoobeAtPhysics

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?

I thought all joints were assumed to be in tension unless they are negative

Last edited: Nov 20, 2013
9. Nov 20, 2013

Pythagorean

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. Nov 20, 2013

NoobeAtPhysics

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. Nov 20, 2013

Pythagorean

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. Nov 20, 2013

NoobeAtPhysics

If Fec is negative then

Fde should theoretically be positive to counteract the y force?

edit:
Oh right!

13. Nov 21, 2013

NoobeAtPhysics

wait still, if Fde must be positive to counteract the y force, why is it pointing downwards?

14. Nov 21, 2013

Pythagorean

who said Fde must be up?