A removable floor safe stands on four identical rectangular feet, resting on a wooden floor tile in an office. The safe has a total weight of 6000 N, shared equally between the four feet. Each foot has a rectangular base measuring 40 mm by 50 mm. The manufacturer states that the…

Answer: ((a)) 0.008 m^2 (i.e. 8.0×10^-3 m^2) ((b)) 7.5×10^5 Pa (i.e. 750 000 Pa) ((c)) The floor tile will crack, because the pressure from the safe's feet (7.5×10^5 Pa) is greater than the maximum pressure the tile can withstand (5.0×10^5 Pa). ((d)) 1.2×10^4 Pa (i.e. 12 000 Pa). The safe's weight (the force) has not changed, but it is now spread over a much larger area, so since p = F/A, the same force over a much bigger area gives a much smaller pressure, well below the tile's maximum rating, so it does not crack.. Part (a): Total contact area of the four feet Find the area of one foot, then multiply by four for all four feet: Aone foot = 40 mm × 50 mm = 2000 mm^2 Atotal = 4 × 2000 mm^2 = 8000 mm^2 Now convert this area into square metres. Since 1 m…

IGCSE Physics 0625: Pressure, Contact Area and a Floor Safe

A removable floor safe stands on four identical rectangular feet, resting on a wooden floor tile in an office. The safe has a total weight of $6000\text{ N}$ , shared equally between the four feet. Each foot has a rectangular base measuring $40\text{ mm}$ by $50\text{ mm}$ . The manufacturer states that the floor tile will crack if the pressure on it ever exceeds $5.0\times10^{5}\text{ Pa}$ .

(a) Calculate the total area of the four feet in contact with the floor, in $\text{m}^2$ . [2]

(b) Calculate the pressure the safe exerts on the floor through its feet. [2]

(c) By comparing your answer to (b) with the maximum pressure the tile can withstand, state whether the floor tile will crack. [2]

(d) To move the safe without damaging the floor, workers place it on a rigid wooden board with an area of $0.50\text{ m}^2$ , so its weight is spread evenly over the whole area of the board. Calculate the new pressure on the floor, and explain why this prevents the tile from cracking. [3]

Part (a): Total contact area of the four feet

Find the area of one foot, then multiply by four for all four feet:

$A_{\text{one foot}} = 40\text{ mm} \times 50\text{ mm} = 2000\text{ mm}^2$

$A_{\text{total}} = 4 \times 2000\text{ mm}^2 = 8000\text{ mm}^2$

Now convert this area into square metres. Since $1\text{ m} = 1000\text{ mm}$ :

$1\text{ m}^2 = (1000\text{ mm})^2 = 1\,000\,000\text{ mm}^2$

So:

$A_{\text{total}} = \frac{8000}{1\,000\,000}\text{ m}^2 = 0.008\text{ m}^2 = 8.0\times10^{-3}\text{ m}^2$

Part (b): Pressure exerted through the feet

Use $p = F/A$ with the safe’s total weight and the total contact area from (a):

$p = \frac{6000\text{ N}}{0.008\text{ m}^2} = 750\,000\text{ Pa} = 7.5\times10^{5}\text{ Pa}$

Part (c): Will the floor tile crack?

Compare the calculated pressure with the maximum the tile can withstand:

$7.5\times10^{5}\text{ Pa} \; (\text{pressure from the feet}) \quad \text{vs} \quad 5.0\times10^{5}\text{ Pa} \; (\text{maximum the tile can take})$

Since $7.5\times10^{5}\text{ Pa} > 5.0\times10^{5}\text{ Pa}$ , the pressure from the safe’s feet exceeds the tile’s maximum rating, so the floor tile will crack.

Part (d): Spreading the load with a board

The safe’s weight is unchanged at $6000\text{ N}$ , but it now acts over the much larger area of the board. Using $p = F/A$ again:

$p = \frac{6000\text{ N}}{0.50\text{ m}^2} = 12\,000\text{ Pa} = 1.2\times10^{4}\text{ Pa}$

Consistency check: the board’s area is $0.50 \div 0.008 = 62.5$ times bigger than the feet’s combined area, so the pressure should be $62.5$ times smaller: $750\,000\text{ Pa} \div 62.5 = 12\,000\text{ Pa}$ . This matches.

This new pressure, $1.2\times10^{4}\text{ Pa}$ , is far below the tile’s maximum rating of $5.0\times10^{5}\text{ Pa}$ . The board does not change the force pressing down (the safe’s weight is exactly the same as before). It only increases the area over which that force acts. Since $p = F/A$ , spreading the same force over a much larger area produces a much smaller pressure, which is why the board stops the tile from cracking.

Final answers

(a) Total contact area of the four feet $=$ $0.008\text{ m}^2$
(b) Pressure through the feet $=$ $7.5\times10^{5}\text{ Pa}$
(c) The tile will crack, since $7.5\times10^{5}\text{ Pa} > 5.0\times10^{5}\text{ Pa}$
(d) Pressure through the board $=$ $1.2\times10^{4}\text{ Pa}$ . Spreading the same force over a much larger area gives a much smaller pressure, well within the tile’s rating