IDNLearn.com provides a comprehensive platform for finding accurate answers. Ask any question and get a detailed, reliable answer from our community of experts.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm Pa}[/tex] and [tex]8.0 \times 10^{3}\; {\rm Pa}[/tex], respectively (assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].)
The [tex]55.5\; {\rm kg}[/tex] box exerts more pressure than the [tex]65.5\; {\rm kg}[/tex] box.
Explanation:
Let [tex]m[/tex] denote the mass of the [tex]55.5\; {\rm kg}[/tex] box.
The weight of that box would be: [tex]W = m\, g[/tex].
The normal force [tex]N[/tex] that this box exerts on the ground would be the same as its weight: [tex]N = W = m\, g[/tex].
If the contact area between the box and the ground is [tex]A[/tex], the pressure [tex]P[/tex] that this box exerts on the ground would be [tex]N / A[/tex]. That is:
[tex]\displaystyle P = \frac{N}{A}= \frac{m\, g}{A}[/tex].
The contact area between the ground and this [tex]55.5\; {\rm kg}[/tex] box is [tex]A = 0.25\; {\rm m} \times 0.20\; {\rm m}[/tex]. Substitute in these values and evaluate to find the pressure:
[tex]\begin{aligned} P &= \frac{N}{A} \\ &= \frac{m\, g}{A} \\ &= \frac{55.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}}}{0.25\; {\rm m} \times 0.20\; {\rm m}} \\ & \approx 1.1 \times 10^{4}\; {\rm Pa}\end{aligned}[/tex].
Similarly, for the [tex]65.5\; {\rm kg}[/tex] box:
[tex]\begin{aligned} P &= \frac{N}{A} \\ &= \frac{m\, g}{A} \\ &= \frac{65.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}}}{0.20\; {\rm m} \times 0.40\; {\rm m}} \\ & \approx 8.0 \times 10^{3}\; {\rm Pa}\end{aligned}[/tex].
Thus, the [tex]55.5\; {\rm kg}[/tex] box exerts greater pressure on the ground than the [tex]65.5\; {\rm kg}[/tex] box.