To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch

Respuesta :

Answer:

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.

Pressure

Friction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.

Pressure exists as force over an area: P=F/A

Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

let's divide P₁ by P₂

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that, we can say:

P₁=(4/0.9025)P₂=4.4P₂ or

P₂=(0.9025/4)P₁=0.23P₁

Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,

To learn more about Pressure refer to:

https://brainly.com/question/912155

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