If an acrobat who weighs 800 N is clinging to a vertical pole using only his hands, neither moving up nor down, can we determine the coefficient of static friction between his hands and the pole? Explain your answer.
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Thank you
To determine the coefficient of static friction between the acrobat's hands and the pole, we can use the fact that he is neither moving up nor down. In this case, the forces acting on the acrobat must be in equilibrium.
Let's consider the forces involved:
1. The weight of the acrobat (800 N) acts vertically downward.
2. The normal force acts vertically upward and is exerted by the pole on the acrobat. Since the acrobat is not moving up or down, the normal force must be equal in magnitude and opposite in direction to the weight (800 N).
Now, the friction force between the hands and the pole acts horizontally and opposes the tendency of the acrobat to slide down. The coefficient of static friction (μs) quantifies the frictional force between two surfaces in contact when there is no relative motion between them.
In order to prevent the acrobat from moving up or down, the static frictional force must be equal in magnitude but opposite in direction to the component of the weight force attempting to pull the acrobat down. This component is given by the weight force multiplied by the sine of the angle between the weight vector and the vertical direction.
Therefore, the frictional force (Ff) can be expressed as:
Ff = μs * Fn,
Where Fn is the normal force and can be assumed to be equal to the weight (800 N) in this case.
Since the acrobat is not sliding up or down, we know that Ff = 0. Hence, we can conclude that the coefficient of static friction (μs) must also be 0.
In summary, based on the fact that the acrobat is not moving up or down while clinging to the pole, we can determine that the coefficient of static friction between his hands and the pole is 0.