A sphere of volume 0.0000013 m³ and density 7800 kg/m³ falls at terminal speed through liquid of density 1200 kg/m³. Take g = 9.81 m/s². Upthrust equals the weight of displaced liquid. (a) Calculate weight, upthrust and viscous drag. (b) Explain why weight does not equal drag. (c) State what measurements establish that terminal speed has been reached.

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Weight 0.09947 N; upthrust 0.0153 N; drag 0.08417 N upward. The sphere's mass is its own density multiplied by volume. Thus weight is ρsVg = 7800 × 0.0000013 × 9.81 = 0.09947 N, acting vertically downward. The displaced-liquid mass is ρlV, so upthrust is 1200 × 0.0000013 × 9.81 = 0.0153 N upward. Use the liquid density for this term, not the sphere's density. Terminal speed means zero acceleration and zero resultant force. Therefore drag + upthrust = weight, giving drag = 0.09947 − 0.0153 = 0.08417 N upward. Weight alone does not balance drag because upthrust is a second upward force. Time the sphere across successive equal vertical distances well below its release point; equal times within measurement uncertainty support constant speed. Keep clear of container walls, which can alter the flow.

A sphere falling through liquid 4

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