Question

Hard Resolving Forces question..... I need to find the angle CED, but the problem is a 3D Force vector one. I can't seem to figure out how to get the needed sides to that triangle to determine the components of the tension. I know what the answer is, but I don't know how to get it. Plz help. I posted the pic

Answer 1

I know the force of 800 lb has something to do with figuring it out, but I am always confused about the relationship between the length of the side, and the magnitude of the forcce that runs along it.

Answer 2

The general approach is to get net forces = 0 and net torques = 0.

The force at A plus the force at B must equal -F at the center, and
the symmetry (F in middle of beam) indicates these two will be -F/2 each.

Ignore the triangle to start. The tension at B will be split equally by the two lines going back to the wall.

tension at B: T sinB = (F/2)

get sinB by extrapolating BE back to the wall: tanB =5/10.

Tension in DE and CE lines will have to have the sum of their components in the direction defined by BE equal to T in BE.
Get the bisector of angle E by extending BE back to the wall, making it the hypotenuse of the triangle with the beam and the wall the other two legs.

I have not done the problem, but I think this will get you there.

Answer 3

I will reread this and take it step by step, but could you explai the relationship between the length of a side of a triangle to the magnitude of the force that runs along it? I don't even get that basic concept....:(

Answer 4

The lengths are useful for getting the angles.

If I pull a string with tension T on an object on the ground, say,
and if the angle between the rope and the ground is B
then the horizontal component of force on the object
is T cosB
and the vertical component is T sinB.

Answer 5

Okay, that is a good explanation, thank you. For my future knowledge, The solution shows what I drew, how did they get that angle?