here, we’ll show you how to calculate the area of a force being applied when we are given the...

Here, we’ll show you how to calculate the area of a force being applied when we are given the force and the pressure.

Calculations Involving Pressure

Example 2

In the previous problem, we discovered that a person with a 55 kg mass, has a force of gravity of 539 Newtons.

P= FA

= 539 N0.028 m2

= 19 250 N/m2Pa

=19.25 N/m2kPa

PP

55 kg539 N

The sole of the boot she is walking on has an area of 0.028 square meters.

P= FA

= 539 N0.028 m2

= 19 250 N/m2Pa

=19.25 N/m2kPa

PP

55 kg539 N

The sole of a boot has a total area of 0.028 m2.

So the pressure applied by the boot is equal to force divided by area

P= FA

= 539 N0.028 m2

= 19 250 N/m2Pa

=19.25 N/m2kPa

PP

55 kg539 N


= FA

= 539 N0.028 m2

= 19 2502 PaP

P

Which is 539 N divided by 0.028 square meters

P= FA

= 539 N0.028 m2

= 19 250 N/m2Pa

=19.25 N/m2kPa

PP

55 kg539 N


= FA

= 539 N0.028 m2

= 19 2502 PaP

P

Which comes out to 19 250 Pascals

P= FA

= 539 N0.028 m2

= 19 250 N/m2Pa

=19.25 N/m2kPa

PP

55 kg539 N


= FA

= 539 N0.028 m2

= 19 2502 PaP

P

In this problem, we’ll consider the same 55 kg person, with a gravitational force of 539 N taking a step with one foot on a snowshoe this time instead of a boot.

55 kg

539 N

A 55 kg person with a gravitational force of 539 N takes a step with one foot that is wearing a snowshoe.

The pressure applied to the snow during this step is found to be 4492 Pascals, considerably less than the 19 250 Pascals applied by the boot.

55 kg

539 N

P = 4492 Pa = 4492 N/m2

A 55 kg person with a gravitational force of 539 N takes a step with one foot that is wearing a snowshoe.The pressure applied to the snow in this case is4492 Pa.

A snowshoe applies less pressure for the same force of gravity because the force is distributed over a larger area than the sole a boot. We’re asked for the area of contact the snowshoe has with the snow.

55 kg

539 N

P = 4492 Pa = 4492 N/m2

A 55 kg person with a gravitational force of 539 N takes a step with one foot that is wearing a snowshoe.The pressure applied to the snow in this case is4492 Pa. What is the area of the snowshoe in m2?

A = ?

Remember that pressure, expressed as 4492 Pascals,

539 N

P = 4492 Pa = 4492 N/m2

A person with a gravitational force of 539 N takes a step with a snowshoe. The pressure applied is 4492 Pa. What is the area of the snowshoe in m2?

A = ?

is the same as 4492 Newtons per square meter.

539 N

P = 4492 Pa = 4492 N/m2


A = ?

The formula for pressure is pressure equals force divided by area, or P equals F divided by A.

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = F

A

Were asked to find the area of the snowshoe so we rearrange this equation to solve for area, A

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = F

A Solve for Area

We start by multiplying both sides of the equation by A (click)

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = F

AA A× ×

P = FA ×

= FP

A

A is in the numerator and denominator on the right side, so it can be cancelled out, (click) leaving us with only F on the right side.

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = F

AA A× ×

P = FA ×

= FP

A

So we’re left with the equation A times P equals F

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = FA ×

= FP

A

Solve for Area

In order to solve this equation for A, we divide both sides of the equation by P (click)

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = FA ×

P

Solve for AreaFPP

A

P is on in the numerator and denominator on the left side, so it can be cancelled, (click), leaving us with only A on the left side.

539 N

P = 4492 Pa = 4492 N/m2


A = ?P = F× F

PP

= FA FP

We’ll just move the equation over here. (click)

539 N

P = 4492 Pa = 4492 N/m2


A = ?

In order to solve for A We’ll start by substituting 539 Newtons in for force.

539 N

P = 4492 Pa = 4492 N/m2


A = ?= FA F

P

=539 N4492 N/m2

=0.12 m2

Then we’ll substitute 4492 Newtons per square meter in for pressure

539 N

P = 4492 Pa = 4492 N/m2


A = ?= FA F

P

=539 N

=0.12 m2

4492 N/m2

539 divided by 4492 is equal to 0.12


= FA FP

=539 N

=0.12 m2

4492 N/m2

Looking at the units, Newtons are divided by the fraction Newtons per square meter


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

N m2/ N= ×

=

In order to divide by a fraction we invert the divisor and multiply,


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

N m2/ N×

Invert the divisor and

multiply

=

giving us Newtons times square meters over Newtons


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

N m2/ N×

Invert the divisor and

multiply

=N

Newtons will cancel out (click)


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

m2/ N×m2

=N

Leaving us with square meters as a unit


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

×m2

=N

Which we’ll write down here


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

×m2

=N

So the final answer is


= FA FP

=539 N

=0.12 m2

4492 N/m2

NN/m2

×m2

the area of the snowshoe is 0.12 square meters


= FA FP

=539 N

=0.12 m2

4492 N/m2

A

A = ?

The area of the snowshoe is 0.12

m2

539 N

Boot Snowshoe

Mass of person 55 kg 55 kgForce of Gravity 539 N 539 N

Area of Contact 0.028 m2 0.12 m2

Pressure 19 250 Pa 4492 Pa

In this table, we’ll compare the mass and force of gravity of the person, the area of contact, and the pressure applied by taking a step with a boot and with a snowshoe

A person with a gravitational force of 539 N takes a step with a snowshoe. The pressure applied is 4492 Pa. The area of the snowshoe is 0.12 m2.

Boot Snowshoe




We’ll start with the boot


Boot Snowshoe




The person wearing the boot has a mass of 55 kg


Boot Snowshoe




Which means she has a force of gravity of 539 Newtons acting on her.


Boot Snowshoe




The area of contact of the boot is 0.028 square meters.


Boot Snowshoe




And the pressure applied by the boot onto the snow is 19 250 Pascals. Because this pressure is so high, the boot will likely sink into the snow.


Boot Snowshoe




Now, we’ll consider these quantities when she takes a step with a snowshoe.


Boot Snowshoe




Her mass is still 55 kg. Here, we’ll ignore any difference in mass that would result by putting on the snowshoe.


Boot Snowshoe




The force of gravity acting on her is still 539 Newtons.


Boot Snowshoe




The area of contact of the snowshoe is 0.12 square meters. The snowshoe distributes the force over an area greater than 4 times than the boot.


Boot Snowshoe




Finally, the pressure the snowshoe applies to the snow is only 4492 Pascals. This is less that a quarter of the pressure applied by the boot.


Boot Snowshoe




This low pressure applied to the snow means the snowshoe will not sink into the snow as much as the boot did.


To summarize, we saw that the boot has a smaller area of contact


539 N

539 N

Smaller area of contact

Greater pressure Will sink into the

snow

Which results in a greater pressure


539 N

539 N



snow

So the boot will likely sink into the snow


539 N

539 N



snow

The snowshoe has a greater area of contact


539 N

539 N



snow

Greater area of contact Lower pressure Will not sink into the

snow

Which results in a lower pressure


539 N

539 N



snow


snow

Therefore, the snowshoe will not sink into the snow as much as the boot did. This is how snowshoes work.


539 N

539 N



snow


snow

here, we’ll show you how to calculate the area of a force being applied when we are given the...

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