- #1

LucasSG

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## Homework Statement

A telescope mirror housing is supported by 6 bars in form of a space truss with geometry defined by the figure below. The total mass is 3000[kg] with center of mass being point G. The distance between:

z axis and points A, B and C is 1 [m]

z axis and points D, E and F is 2.5[m]

Using the exact same coordinate system the problem gives, the points coordinates are:

G=(0,0,1)

A=(-1/2, √3/2, 0)

B=(-1/2, -√3/2, 0)

C=(1,0,0)

D=(-5/2, 0, -4)

E=(5/4, (-5√3)/4, -4)

F=(5/4, (5√3)/4, -4)

The force vector P is equal to:

P=(30*cos(20)cos(60), -30cos(20)sin(60), -30sin(20))[kN] (using g=10m/s²)(Ball and socket supports)

## Homework Equations

∑Fx=0 (1)

∑Fy=0 (2)

∑Fz=0 (3)

∑Mx=0 (4)

∑My=0 (5)

∑Mz=0 (6)

## The Attempt at a Solution

First thing i thought was to consider A, B and C "ball and socket supports" detaching them from the bars. So i would have reactions on x, y and z directions in each of these three points. Then i would consider the entire part between G and the points A, B and C to be a rigid body. The forces that act on each point would be (using the same coordinate system the problem gives me):

Fa=(Fax, Fay, Faz) [kN]

Fb=(Fbx, Fby, Fbz) [kN]

Fc=(Fcx, Fcy, Fcz) [kN]

P=(30*cos(20)cos(60), -30cos(20)sin(60), -30sin(20))[kN]

Then i used ∑M=0 on point A, so i got:

∑Ma= AB x Fb +Ac x Fc + AG x P = 0

= (-√3Fbz, 0, √3Fbx) + (-0.86Fcz, -1.5Fcz, 0.86Fcx + 1.5Fcy) + (15√3*(cos(20)+sin(20)), 15cos(20), 15sin(20)) And that's where I'm stuck. Using equations 1-6 now i got 6 equations and 9 unknowns, i guess now i need to find symmetry between the forces on those 3 points, but i can't seem to find any new information about those forces that can help me get the other 3 equations i need.

I suppose finding the sum of the moments on points B or C won't give me any new information.

Am i trying to solve this problem the right way or there's a better way of finding the reactions on A, B and C to analyze the truss? If so, what can i do next? I'm really stuck right now...

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