The cables AB and AC and AD support a crate as shown below D

The cables AB and AC and AD support a crate as shown below. Determine the maximum weight of the crate so that the tension developed in any cables does not exceed 700 Ib. Determine the tensions in three cables supporting the 3000 Ib slab.

Solution

The weight acts only in –z direction

w^ = -wk

AD has only x-component

let i,j,k be the unit vectors along x,y,z directions

length of AB = sqrt(1² + 4² +3² )= 26

length of AD = 5

length of AC = sqrt(4² + 3² + 2² )= 29

unit vector along AB = b^ = (-4i +j+3k)/ 26

unit vector along AC = c^ = (-4i -3j+2k)/ 29

unit vector along AD = d^{^}   = i

let Td, Tb and Tc be the magnitude of tension in the cables AD, AB and AC

then Td^ = Td i

         Tb^{^} = Tb(-4i +j+3k)/ 26

Tc^ =Tc(-4i -3j+2k)/ 29

resolving the components along the x,y, z directions and balancing the forces

w = 3Tb/26 + 2Tc/29 -(1)   --- z-axis

Td = -4Tb/26 -4Tc/29 -(2)   ----x-axis

Tb/26 = 3Tc/29           --(3) ---y-axis

we have a general condition Ta<700, Tb<700,Tc<700

Tb = 2.84 Tc

Td = 0.78Tb + 0.74 Tc = 2.96Tc

w= 0.59Tb + 0.37Tc = 2.05Tc

from the above equations we see Td has the maximum tension and must not be >700

pegging Td at 700

Td = 2.96Tc = 700

Tc = 236.45

w= 2.05*236.45 = 484.78 lb