A ball is falling at terminal speed in still air. The forces acting on the ball are upthrust,

Question 1060: [Dynamics > Resultant force] (Past Exam Paper – June 2008 Paper 1 Q12 & November 2014 Paper 11 & 12 Q11)

A ball is falling at terminal speed in still air. The forces acting on the ball are upthrust, viscous drag and weight.
What is the order of increasing magnitude of these three forces?
A upthrust → viscous drag → weight
B viscous drag → upthrust → weight
C viscous drag → weight → upthrust
D weight → upthrust → viscous drag



Solution 1060:
Answer: A.
Sketching a force diagram would have greatly helped in answering this question. The weight is downward, while the viscous drag and the upthrust would be upwards, opposing the weight.

At terminal speed, the resultant force on the ball is zero. As weight is the only downward force, it must be the largest of the three.

The upthrust is a result of the force due to the difference in pressure at the top and bottom of the ball.

Pressure P = hρg         

and Force = Pressure / Area

Air is not very dense (ρ is small) and the difference in height, h, from the top of the ball to its bottom is also very small. Thus, the upthrust force is small.

Many students would think that the Archimedean upthrust on a body in air is quite large. This is a mistake. The reverse is the case. 

The upthrust will only be approximately a thousandth or less than the weight of a solid.

Viscous drag (air resistance), on the other hand, will be almost as large as the weight when a body is at terminal velocity.

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