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The head of a thumbtack has an area of 500 mm² - Leaving Cert Physics - Question b - 2008

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The head of a thumbtack has an area of 500 mm². Its point has an area of 0.3 mm². The pressure exerted at the head of the thumbtack is 12 Pa. What is the pressure ex... show full transcript

Worked Solution & Example Answer:The head of a thumbtack has an area of 500 mm² - Leaving Cert Physics - Question b - 2008

Step 1

At head:

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Answer

To find the force exerted at the head of the thumbtack, we can use the formula for pressure:

F=P×AF = P \times A

Here, the pressure (P) at the head is 12 Pa, and the area (A) of the head is 500 mm²:

F=12 Pa×(500×10−6 m2)F = 12 \text{ Pa} \times (500 \times 10^{-6} \, \text{m}^2)

Calculating this gives:

F=12×500×10−6=6.0×10−3 NF = 12 \times 500 \times 10^{-6} = 6.0 \times 10^{-3} \text{ N}

Step 2

At point:

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Answer

Now, we need to calculate the pressure exerted at the point of the thumbtack using the same force we calculated before. The area of the point is 0.3 mm²:

Using the pressure formula again:

P=FAP = \frac{F}{A}

Substituting the values gives:

P=6.0×10−3 N(0.3×10−6 m2)P = \frac{6.0 \times 10^{-3} \text{ N}}{(0.3 \times 10^{-6} \, \text{m}^2)}

Calculating the pressure:

P=6.0×10−30.3×10−6=2.0×104 PaP = \frac{6.0 \times 10^{-3}}{0.3 \times 10^{-6}} = 2.0 \times 10^{4} \text{ Pa}

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