**Problem Context:**John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high. The handles make an angle of θ = 16.0° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 411 N is exerted at the center of the wheel, which has a radius of 18.5 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.)**Image Description:**The image shows a man pushing a wheelbarrow with a child inside. The wheelbarrow's handles are at an angle θ with the ground, leading to a force directed toward the center of the wheel. A brick is stopping the wheelbarrow.**Part (a):**(a) What force (in N) must John apply along the handles to just start the wheel over the brick?- Input box for the answer with an error message: "differs from the correct answer by more than 10%. Double check your calculations. N"**Part (b):**(b) What is the force (magnitude in kN and direction in degrees clockwise from the −x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick?- Input boxes for: - Magnitude: ____ kN - Direction: ____° clockwise from the −x-axis**Educational Explanation:**This problem highlights the application of physics principles related to forces, angles, and rotations. Understanding how forces are distributed and affect movement, specifically in overcoming obstacles, is crucial in real-world applications. Students are encouraged to think about the balance of forces and the influence of angles on mechanical advantage.

Given-Information:- H = 8 cmθ = 16°Downward Force = 411 NRadius,r =18.5 cm To-Determine:- a. Force john must apply along the handles to just start the wheel over the brick.b. Force that brick exerts on wheel just as the wheel begins to lift over the brick. a. The vertical distance from the corner of the brick to the center of the wheel,h = 18.5 - 8 h = 10.5 cm The horizontal distance from the corner of the brick to the center of the wheel is,d = 18.52-10.52d = 15.23 cm Taking sum of moments about the corner of brick,411+Fsin16°×15.23 = Fcos16×10.56259.53 + F4.1979 = 10.093F6259.53=10.093F-4.1979F6259.53=5.895FF=1061.837 Nb. R = 411+1061.837sin162+1061.837cos162R = 1239.759 NR = 1240 Nθ = tan-1411+1061.837sin161061.837cos16θ = tan-1703.681020.70θ = 34.58°a. Force john must apply along the handles to just start the wheel over the brick, F = 1061.837Nb. Force that brick exerts on wheel just as the wheel begins to lift over the brick, R = 1240 N, θ=34.58°

Answered: er of the wheel, which has a radius of…

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Mechanical Engineering

er of the wheel, which has a radius of 18.5 cm. Assume the brick remains fix force applied by John is directed exactly toward the center of the wheel. (C

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.73P

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