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A hot air balloon has a volume of 1.2 m³ and a fabric mass of 0.3 kg. If the density of air is 1.25 kg/m³ and the density of helium is 0.018 kg/m³, calculate the maximum tension on the string.

Physics

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ii) I cannot determine the average relative density of the spirit as the necessary information (such as its density, mass, or volume) is not provided in the visible part of the question. This question likely refers to data from a preceding section.

c)
Given:
Volume of balloon, V=1.2 m3V = 1.2 \text{ m}^3
Mass of fabric, mfabric=0.3 kgm_{fabric} = 0.3 \text{ kg}
Density of air, ρair=1.25 kg/m3\rho_{air} = 1.25 \text{ kg/m}^3
Density of helium, ρhelium=0.018 kg/m3\rho_{helium} = 0.018 \text{ kg/m}^3
Acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2 (assumed standard value)

The maximum tension on the string occurs when the balloon is fully inflated and the net upward force (buoyant force) is balanced by the total downward forces (weight of helium, weight of fabric, and tension in the string).

  1. Calculate the buoyant force (FBF_B) acting on the balloon.

    The buoyant force is equal to the weight of the air displaced by the balloon.

    FB=ρair×V×gF_B = \rho_{air} \times V \times g FB=(1.25 kg/m3)×(1.2 m3)×(9.8 m/s2)F_B = (1.25 \text{ kg/m}^3) \times (1.2 \text{ m}^3) \times (9.8 \text{ m/s}^2) FB=14.7 NF_B = 14.7 \text{ N}
  2. Calculate the weight of the helium gas (WheliumW_{helium}) inside the balloon.

    First, find the mass of the helium: mhelium=ρhelium×Vm_{helium} = \rho_{helium} \times V.

    mhelium=(0.018 kg/m3)×(1.2 m3)m_{helium} = (0.018 \text{ kg/m}^3) \times (1.2 \text{ m}^3) mhelium=0.0216 kgm_{helium} = 0.0216 \text{ kg}

    Now, calculate its weight:

    Whelium=mhelium×gW_{helium} = m_{helium} \times g Whelium=(0.0216 kg)×(9.8 m/s2)W_{helium} = (0.0216 \text{ kg}) \times (9.8 \text{ m/s}^2) Whelium=0.21168 NW_{helium} = 0.21168 \text{ N}
  3. Calculate the weight of the balloon fabric (WfabricW_{fabric}).

    Wfabric=mfabric×gW_{fabric} = m_{fabric} \times g Wfabric=(0.3 kg)×(9.8 m/s2)W_{fabric} = (0.3 \text{ kg}) \times (9.8 \text{ m/s}^2) Wfabric=2.94 NW_{fabric} = 2.94 \text{ N}
  4. Determine the maximum tension (TT) on the string.

    The forces are in equilibrium, so the upward force equals the sum of downward forces:

    FB=Whelium+Wfabric+TF_B = W_{helium} + W_{fabric} + T

    Rearrange to solve for TT:

    T=FB−Whelium−WfabricT = F_B - W_{helium} - W_{fabric} T=14.7 N−0.21168 N−2.94 NT = 14.7 \text{ N} - 0.21168 \text{ N} - 2.94 \text{ N} T=11.54832 NT = 11.54832 \text{ N}

    Rounding to three significant figures:

    T≈11.5 NT \approx 11.5 \text{ N}

    The maximum tension on the string is 11.5 N\boxed{\text{11.5 N}}.

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