Question

plz answer this quest....​

Answer 1

Answer:

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Explanation:

using symmetry here we see that all forces

f1 = f2 = f3 = kQq / r power 2

here r = AO = BO = CO

using trigonometry we can find "r"

here we can see that length of each side of triangle ABC is " l "

=> in triangle BOD

angle OBD = 30 degree

angle BD = L /2

cos 30 = L/2r => r = L / 2 into sec 30 degree

=> r = L/2 into 2/√3 = ( 1/√3) L ----------1

so we can that all forces are equal

F1 = F2 = F3 = kQq / ( L / √3 ) power 2 = 3 kQq / L power 2

and angle between each force is 120 degree

using Lami's theorem all forces are equal and magnituide and angle between them is also equal and planar then resultant force is 0

=> fnet = f1 + f2 + f3 = 0

here we add forces using vectors to get net force = 0

so, net force on charge " Q " placed at centroid is zero

Answer 2

Answer:

Mass of car m=1800kg

Distance between front and back axles d=1.8m 

Distance between center of gravity and front axle=1.05m

Let Rb and Rf be the forces exerted by the level ground on the back and front wheels respectively.

At translational equilibrium:

Rf+Rb=mg=17640 N.....(i)

For rotational equilibrium, on taking the torque about the C.G., we have:

Rf(1.05)=Rb(1.8−1.05)

⟹Rb=1.4Rf.......(ii)

Solving (i) and (ii) gives

Rf=7350 N

Rb=10290 N

The force exerted on each front wheel =7350/2=3675 N

The force exerted on each back wheel =10290/2=5145N