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Why is the highlighted portion not negative? The car is accelerating to the left so it should make a counterclockwise rotation on point A right

Why is the highlighted portion not negative? The car is accelerating to the left so it should make a counterclockwise rotation on point A right

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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From Hibbeler's Engineerin Mechanics Dynamics Textbook

Why is the highlighted portion not negative? The car is accelerating to the left so it should make a counterclockwise rotation on point A right?

The car shown in Fig. 17–10a has a mass of 2 Mg and a center of mass at G. Determine the acceleration if the rear “driving" wheels are always slipping, whereas the front wheels are free to rotate. Neglect the mass 0.3 m of the wheels. The coefficient of kinetic friction between the wheels and 1.25 m '0.75 m the road is µz = 0.25. (a) SOLUTION I Free-Body Diagram. As shown in Fig. 17-10b, the rear-wheel frictional force Fg pushes the car forward, and since slipping occurs, FB = 0.25N.. The frictional forces acting on the front wheels are zero, since these wheels have negligible mass.* There are three unknowns in the problem, NA, Ng, and aç. Here we will sum moments about the mass center. The car (point G) accelerates to the left, i.e., in the negative x direction, Fig. 17–10b. aG 2000 (9.81) N 0.3 m Equations of Motion. F = 0.25 Ng A NA |-1.25 m- 4ΣF, m (aς).; -0.25NB = -(2000 kg)ac (1) 0.75 m +1£F, = m(ac),; NA + Ng – 2000(9.81) N = 0 (2) 6+EMG = 0; -NĄ(1.25 m) – 0.25Ng(0.3 m) + Ng(0.75 m) = 0 (3) (b) Solving, ag 1.59 m/s? + Ans. 2000 (9.81) N NA = 6.88 kN 0.3 m Ng = 12.7 kN - FB = 0.25 Ng SOLUTION II Free-Body and Kinetic Diagrams. If the "moment" equation is applied about point A, then the unknown NA will be eliminated from the equation. To "visualize" the moment of maç about A, we will include the kinetic diagram as part of the analysis, Fig. 17–10c. Equation of Motion. |-1.25 m-- 0,75 m || 2000 aG 6+EMa = £(M)a; Ng(2 m) – [2000(9.81) NJ(1.25 m) = 0.3 m (2000 kg)ac(0.3 m) (c) Solving this and Eq. 1 for aç leads to a simpler solution than that obtained from Eqs. 1 to 3. Fig. 17-10 *With negligible wheel mass, la = 0 and the frictional force at A required to turn the wheel is zero. If the wheels' mass were included, then the solution would be more involved, since a general-plane-motion analysis of the wheels would have to be considered (see Sec. 17.5).
Transcribed Image Text:The car shown in Fig. 17–10a has a mass of 2 Mg and a center of mass at G. Determine the acceleration if the rear “driving" wheels are always slipping, whereas the front wheels are free to rotate. Neglect the mass 0.3 m of the wheels. The coefficient of kinetic friction between the wheels and 1.25 m '0.75 m the road is µz = 0.25. (a) SOLUTION I Free-Body Diagram. As shown in Fig. 17-10b, the rear-wheel frictional force Fg pushes the car forward, and since slipping occurs, FB = 0.25N.. The frictional forces acting on the front wheels are zero, since these wheels have negligible mass.* There are three unknowns in the problem, NA, Ng, and aç. Here we will sum moments about the mass center. The car (point G) accelerates to the left, i.e., in the negative x direction, Fig. 17–10b. aG 2000 (9.81) N 0.3 m Equations of Motion. F = 0.25 Ng A NA |-1.25 m- 4ΣF, m (aς).; -0.25NB = -(2000 kg)ac (1) 0.75 m +1£F, = m(ac),; NA + Ng – 2000(9.81) N = 0 (2) 6+EMG = 0; -NĄ(1.25 m) – 0.25Ng(0.3 m) + Ng(0.75 m) = 0 (3) (b) Solving, ag 1.59 m/s? + Ans. 2000 (9.81) N NA = 6.88 kN 0.3 m Ng = 12.7 kN - FB = 0.25 Ng SOLUTION II Free-Body and Kinetic Diagrams. If the "moment" equation is applied about point A, then the unknown NA will be eliminated from the equation. To "visualize" the moment of maç about A, we will include the kinetic diagram as part of the analysis, Fig. 17–10c. Equation of Motion. |-1.25 m-- 0,75 m || 2000 aG 6+EMa = £(M)a; Ng(2 m) – [2000(9.81) NJ(1.25 m) = 0.3 m (2000 kg)ac(0.3 m) (c) Solving this and Eq. 1 for aç leads to a simpler solution than that obtained from Eqs. 1 to 3. Fig. 17-10 *With negligible wheel mass, la = 0 and the frictional force at A required to turn the wheel is zero. If the wheels' mass were included, then the solution would be more involved, since a general-plane-motion analysis of the wheels would have to be considered (see Sec. 17.5).
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