Since the equation that should be used is sum of F in the y direction equals to
ID: 2303632 • Letter: S
Question
Since the equation that should be used is sum of F in the y direction equals to m multplied bu ag Why it is not like that when i substitute: Na+Nb-2000(kg)(9.8)= 2000*9.8 Why its equal to zero in the right hand side ? 17.3 EQuATIONS OF MoTION: TRANLATON 415 EXAMPLE 17.5 17-10a has a mass of 2 Mgand a center of in Fi leration if the rear "driving" wheels are whereas the front wheels are tree to rotate. Neglect ajways sn the wheels. The coefficient of kinetic friction b Neglect s The coefficient of kinetic friction between the at G. Determine t 0.25 .25m 075 ls and the road is A SOLUTION ar am. As shown in Fig, 17-10b, the rear-wheel c F pushes the car forward, and since slipping occurs, 1 %. The frictional forces acting on the front wheels are zero, 2s wheels have negligible mass. There are three unknowns in Fs since thesN.. N, and ag. Here we will sum moments about the mass te o he car (point G) accelerates to the left, i.e, in the negative 11-1 2000 (9.81) N Fig. Equations of Motion. 0.25N-(2000 kg)ag (1)A F-025 N Na NAL NA + NB-2000(9.81 ) N = 0 m) + Ne(0.75 m) 0 (2) (3) -125m a75 m aG 1.59 m/s- NA 6.88 kN Ng -12.7 kN Ans. 2000 (9.81) N 0.3m SOLUTION II Free-Body and Kinetic Diagrams. If the "moment" equation is A applied about point A, then the unknown NA will be eliminated from the N, equation. To "visualize" the moment of mac about A, we will include the 1.25m kinetic diagram as part of the analysis, Fig. 17-10c. F-0.25 N Equation of Motion. +2MA-(M)A: Ne(2 m) 2000(9.81) NJ(1.25 m) - 03m (2000 kg)ad(0.3 m) e) Salving this and Eq. 1 for ag leads to a simpler solution than that otained from Eqs. I' to 3. Fig. 17-10 ith negligible wheel mass, /?-o and the frictional force at A required to turn the heelis era Ifthe wheels' snce a Sec 175) mass were included, then the solution would be more involved, lane-motion analysis of the wheels would have to be considered (see general-plExplanation / Answer
I got your point and that's a very genuine query in force and moment resolution.
See, for any body which is stable under all physical condition, the forces and moments in that body will always be in balanced condition.
For example, if a bar of 20 N is put on two supports then to balance the load of 29 N, support will produce a reaction of 10 N and 10 N in upward direction.
Same way moment will also balance and conserve.
In your case, if we consider y direction, so reactions Na and Nb acting upward, to counter them force equivalent to weight of car will act downwards.
I hope you got my point. In case any doubt, feel free to ask, would be happy to help you.
Comment in case any doubt, please rate my answer....