Question #74469

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Question #74469

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Answer 1 · 2016-04-21T00:07:18

$2.5 \times 10^{- 3} m^{- 2}$ $2.5xx10^-3m^-2$

Explanation:

The cylinder with mass $2 k g and$ $2kg and$ area of cross section $5 c m^{2}$ $5cm^2$ is just able to be attached to the ceiling with the adhesive. Changed to SI units for sake of conformity.

The adhesive is able to balance the weight of the cylinder $m g$ $mg$ , where $g$ $g$ is acceleration due to gravity.

It implies that the adhesive has just capacity of holding downwards force per unit area $= \frac{Weight}{Area of cross section}$ $="Weight"/"Area of cross section"$
$= \frac{2 g}{5 \times 10^{- 4}} N m^{- 2}$ $=(2g)/(5xx10^-4)Nm^-2$
When the mass if the cylinder is changed to $10 k g$ $10kg$ .
We see that the mass has increased by $\frac{10}{2} = 5 times$ $10/2=5 " times"$

As the holding power of the adhesive does not change, therefore the area of cross section must be increased by the same factor for the new cylinder to stay attached with the ceiling.
Therefore, required area of cross section $= 5 \times 10^{- 4} \times 5$ $=5xx10^-4xx5$
$= 2.5 \times 10^{- 3} m^{- 2}$ $=2.5xx10^-3m^-2$

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Question #74469

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