Abox is held at rest by two ropes that form 30° angles with the vertical. the tension t in either rope is 42 n. what is the weight (w) of the box?

So I'm picturing the ropes as both tied to the same point on the top of
the box, and forming a " V " as they leave it, with each side of the 'V'
tilted 30° away from the vertical.  It really doesn't matter whether they're
tied to the same point or to different points.  Everything is in equilibrium,
nothing is moving, we don't have to worry about the horizontal components
of the tension, and all we have to find is the vertical components.

The vertical component of tension in each rope is  42 cos(30°).

There are 2 ropes, so the net upward force on the box is

                                     84 cos(30°)  =  72.75 newtons   (rounded)

Everything is in equilibrium, nothing is accelerating, so the net
upward force on the box is equal to its weight.

When the rock is suspended in the air, the net force on it is

∑ F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

∑ F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

∑ F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³