Answer:
293 kg
Explanation:
Let's say the tension in each cable is Tb, Tc, and Td.
First, find the length of cable AD:
r = √(2² + 2² + 1²)
r = 3
Using similar triangles:
Tdx = 2/3 Td
Tdy = 2/3 Td
Tdz = 1/3 Td
Sum of the forces in the x direction:
∑F = ma
Tb − 2/3 Td = 0
Td = 3/2 Tb
Sum of the forces in the y direction:
∑F = ma
2/3 Td − Tc = 0
Td = 3/2 Tc
Sum of the forces in the z direction:
∑F = ma
1/3 Td − mg = 0
Td = 3mg
From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.
8600 N = 3mg
m = 8600 N / (3 × 9.8 m/s²)
m ≈ 292.5 kg
Rounded to three significant figures, the maximum mass of the crate is 293 kg.
