Question

1. a 9250 kg bridge going east - west is supported by two pillars. the bridge is 47.0 m long with a pillar 5.0 m from the west end and 10 meters from the east end. how much force is applied on each of the pillars? 7. a 1720 kg truck drives across the same bridge going east to west. how much force is applied on each pillar when the truck is 12 m across?

Explanation:

Step1: Calculate bridge weight

$W_b = m_b g = 9250\ \text{kg} \times 9.81\ \text{m/s}^2 = 90742.5\ \text{N}$

Step2: Define distances for torque

Let Pillar A (west) be pivot. Distance from A to bridge's center: $\frac{47.0}{2} - 5.0 = 18.5\ \text{m}$. Distance from A to Pillar B: $47.0 - 5.0 - 10.0 = 32.0\ \text{m}$.

Step3: Torque equilibrium for $F_B$

Sum of torques about A = 0:
$F_B \times 32.0 = W_b \times 18.5$
$F_B = \frac{90742.5 \times 18.5}{32.0} \approx 52000\ \text{N}$

Step4: Force equilibrium for $F_A$

Sum of vertical forces = 0:
$F_A + F_B = W_b$
$F_A = 90742.5 - 52000 \approx 38700\ \text{N}$

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Step1: Calculate truck weight

$W_t = m_t g = 1720\ \text{kg} \times 9.81\ \text{m/s}^2 = 16873.2\ \text{N}$

Step2: Distance of truck from Pillar A

Truck is 12 m across bridge: distance from A = $12.0 - 5.0 = 7.0\ \text{m}$

Step3: Torque equilibrium for $F_B'$

Sum of torques about A = 0:
$F_B' \times 32.0 = W_b \times 18.5 + W_t \times 7.0$
$F_B' = \frac{90742.5 \times 18.5 + 16873.2 \times 7.0}{32.0} \approx 55700\ \text{N}$

Step4: Force equilibrium for $F_A'$

Sum of vertical forces = 0:
$F_A' + F_B' = W_b + W_t$
$F_A' = 90742.5 + 16873.2 - 55700 \approx 51900\ \text{N}$

Answer:

Problem 6:

Force on west pillar (A): $\approx 3.87 \times 10^4\ \text{N}$
Force on east pillar (B): $\approx 5.20 \times 10^4\ \text{N}$

Problem 7:

Force on west pillar (A): $\approx 5.19 \times 10^4\ \text{N}$
Force on east pillar (B): $\approx 5.57 \times 10^4\ \text{N}$