Let's solve the problem step by step: **Given:** - Weight of vehicle (\( W \)) = 6,100 lbs - Speed (\( v \)) = 33 miles/hour - Radius (\( r \)) = 250 feet - Acceleration due to gravity (\( g \)) = 32 ft/s² We are to find the **frictional (centripetal) force** needed to keep the vehicle moving in a circle, i.e., the force the road must exert on the tires. --- ### **Step 1: Convert speed to feet per second** 1 mile = 5280 feet 1 hour = 3600 seconds \[ v = 33 \text{ mi/hr} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3600 \text{ s}} \] \[ v = 33 \times \frac{5280}{3600} \] \[ v = 33 \times 1.4667 = 48.401 \text{ ft/s} \] --- ### **Step 2: Find mass of the vehicle in slugs** Weight (\( W \)) = mass (\( m \)) × gravity (\( g \)) \[ W = m \cdot g \implies m = \frac{W}{g} \] \[ m = \frac{6100}{32} = 190.625 \text{ slugs} \] --- ### **Step 3: Calculate the required centripetal force** Centripetal force formula: \[ F = m \frac{v^2}{r} \] \[ F = 190.625 \times \frac{(48.401)^2}{250} \] First, calculate \( (48.401)^2 \): \[ (48.401)^2 = 2343.7 \] Now plug in the values: \[ F = 190.625 \times \frac{2343.7}{250} = 190.625 \times 9.3748 = 1,787.3 \text{ lbs} \] --- ### **Step 4: Round to one decimal place** \[ F \approx \boxed{1787.3} \text{ lbs} \] --- ## **Final Answer** **The frictional force that the road surface must exert on the tires is \( \boxed{1787.3} \) lbs.**

Let's carefully work through each step of the problem with detailed calculations to find the frictional force that the road surface must exert on the tires. --- ### **Given Data:** | Quantity | Value | |------------------------------|------------------------------| | Vehicle weight (\(W\)) | 6100 pounds | | Speed (\(v\)) | 33 miles/hour | | Radius of circle (\(r\)) | 250 feet | | Acceleration due to gravity (\(g\)) | 32 ft/sec² | --- ### **Step 1: Convert the vehicle's speed from miles/hour to feet/sec** **Conversion factors:** - 1 mile = 5280 feet - 1 hour = 3600 seconds **Calculation:** \[ v = 33\, \text{miles/hour} \times \frac{5280\, \text{feet}}{1\, \text{mile}} \times \frac{1\, \text{hour}}{3600\, \text{seconds}} \] Simplify the expression: \[ v = 33 \times \frac{5280}{3600} \text{ ft/sec} \] Compute numerator: \[ 33 \times 5280 = 174,240 \] Divide by 3600: \[ v = \frac{174,240}{3600} \approx 48.4\, \text{ft/sec} \] **Result:** \[ \boxed{v \approx 48.4\, \text{ft/sec}} \] --- ### **Step 2: Find the mass of the vehicle in slugs** Since weight \(W = m \times g\), \[ m = \frac{W}{g} \] where: - \(W = 6100\, \text{lbs}\) - \(g = 32\, \text{ft/sec}^2\) **Calculation:** \[ m = \frac{6100}{32} \approx 190.625\, \text{slugs} \] **Result:** \[ \boxed{m \approx 190.6\, \text{slugs}} \] --- ### **Step 3: Calculate the centripetal force** The force needed to keep the vehicle moving in a circle (centripetal force): \[ F = m \frac{v^2}{r} \] Substitute known values: \[ F = 190.625 \times \frac{(48.4)^2}{250} \] First, compute \(v^2\): \[ (48.4)^2 = 2343.36 \] Now, divide by radius: \[ \frac{2343.36}{250} = 9.37344 \] Finally, multiply by mass: \[ F = 190.625 \times 9.37344 \approx 1787.3\, \text{lbs} \] --- ### **Final step: Round to one decimal place** \[ \boxed{\textbf{F} \approx \mathbf{1787.3\, \text{lbs}}} \] --- ## **Final Answer:** **The maximum frictional force (which the road surface must exert on the tires) is approximately \(\boxed{1787.3}\) pounds.**