A 100-kg person sits on a 5-kg bicycle. The total weight is borne equally by the two wheels of the bicycle. The tires are 2.0 cm wide and are inflated to a gauge pressure of $8.0 \times 10 ^ { 5 } \mathrm {~Pa}$ What length of each tire is in contact with the ground?
A) 1.6 cm
B) 2.4 cm
C) 1.8 cm
D) 6.4 cm
E) 3.2 cm

Question

A 100-kg person sits on a 5-kg bicycle. The total weight is borne equally by the two wheels of the bicycle. The tires are  2.0 cm wide and are inflated to a gauge pressure of $8.0 \times 10 ^ { 5 } \mathrm {~Pa}$ What length of each tire is in contact with the ground?
A)  1.6 cm
B)  2.4 cm
C)  1.8 cm
D)  6.4 cm
E)  3.2 cm

Quizplus · Accepted Answer

The total weight of the person and bicycle is $105 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1029 \, \text{N}$. Since the weight is equally distributed, each wheel supports $1029 \, \text{N} / 2 = 514.5 \, \text{N}$. The pressure $P$ is the force $F$ divided by the contact area $A$, and $A$ can be expressed as the product of the length of the tire in contact with the ground ($l$) and its width ($w$). Given $P = 8.0 \times 10^5 \, \text{Pa}$, $F = 514.5 \, \text{N}$, and $w = 0.02 \, \text{m}$, we solve for $l$ in $P = F / A = F / (l \times w)$, yielding $l = F / (P \times w) = 514.5 / (8.0 \times 10^5 \times 0.02) = 3.2 \, \text{cm}$.

A 100-Kg Person Sits on a 5-Kg Bicycle 8.0×105 Pa8.0 \times 10 ^ { 5 } \mathrm {~Pa}8.0×105 Pa

A 100-Kg Person Sits on a 5-Kg Bicycle $8.0 \times 10 ^ { 5 } \mathrm {~Pa}$