### Problem StatementA 1555-kg automobile has a wheelbase (the distance between the axles) of \( \ell = 2.85 \, \text{m} \). The automobile's center of mass is on the centerline at a point \( 1.35 \, \text{m} \) behind the front axle. Find the force exerted by the ground on each wheel.### Part 1 of 3 - ConceptualizeSince the center of mass is located in the front half of the car, there should be more force on the front wheels than the rear ones. The sum of the wheel forces must equal the weight of the car.### Part 2 of 3 - CategorizeUse the figure as a force diagram. Apply Newton's second law and sum the torques to find the unknown forces.**Diagram Description:**- A car illustration with labeled forces: - \( 2F_f \): Force at the front wheels. - \( 2F_r \): Force at the rear wheels. - \( F_g \): Gravitational force acting downward at the car's center of mass.- Dimensions are indicated: - \( d = 1.35 \, \text{m} \) from the front axle to the center of mass. - \( \ell = 2.85 \, \text{m} \) total wheelbase length.### Part 3 of 3 - AnalyzeThe torque about the front axle is zero, so we have\[\sum \tau_f = 0 = (d \cdot F_g) - \ell (2F_r)\]Thus, the force at each rear wheel is given by\[F_r = \frac{d \cdot F_g}{2\ell}\]Attempted Calculations:\[\left(1.35 \, \text{m}\right) \times (1.555 \, \text{kg}) \rightarrow \text{Incorrect calculation, check units and accuracy.}\]Re-evaluating gives:\[= \frac{2 \left(1.35 \right) \, \text{m}}{\left(2.85\right) \, \text{m}} \rightarrow \text{Check calculations.}\]Since the sum of the wheel forces equals the weight of

a) Substituting the numerical values in the formulas given we have,…

A 1 555-kg automobile has a wheel base (the distance between the axles) of l = 2.85 m. The automobile's center of mass is on the centerline at a point 1.35 m behind the front axle. Find the force exerted by the ground on each wheel. Part 1 of 3 - Conceptualize Since the center of mass is located in the front half of the car, there should be more force on the front wheels than the rear ones. The sum of the wheel forces must equal the weight of the car. Part 2 of 3 - Categorize Use the figure as a force diagram. Apply Newton's second law and sum the torques to find the unknown forces. Part 3 of 3 - Analyze The torque about the front axle is zero, so we have Στ = 0 = (d Fg) - l (2F₁). Thus the force at each rear wheel is given by d F F₁ = F₁ = 2Ff 9 28 Fg m)( X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step da -2F₁ 2 2F, N. Since the sum of the wheel forces equals the weight of the car, we have 2F₁ + 2F₁ = Fg. Substitu and solving for F₁, we obtain the force at each front wheel. 2(E X Your response differs from the correct answer by more than 10%. Double check your calculations. m (N)-2( 2 N. 1.35 (N) 1.555 1.35

A 1 555-kg automobile has a wheel base (the distance between the axles) of l = 2.85 m. The automobile's center of mass is on the centerline at a point 1.35 m behind the front axle. Find the force exerted by the ground on each wheel. Part 1 of 3 - Conceptualize Since the center of mass is located in the front half of the car, there should be more force on the front wheels than the rear ones. The sum of the wheel forces must equal the weight of the car. Part 2 of 3 - Categorize Use the figure as a force diagram. Apply Newton's second law and sum the torques to find the unknown forces. Part 3 of 3 - Analyze The torque about the front axle is zero, so we have Στ = 0 = (d Fg) - l (2F₁). Thus the force at each rear wheel is given by d F F₁ = F₁ = 2Ff 9 28 Fg m)( X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step da -2F₁ 2 2F, N. Since the sum of the wheel forces equals the weight of the car, we have 2F₁ + 2F₁ = Fg. Substitu and solving for F₁, we obtain the force at each front wheel. 2(E X Your response differs from the correct answer by more than 10%. Double check your calculations. m (N)-2( 2 N. 1.35 (N) 1.555 1.35

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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