Lets say we have a car with a mass of 1000kg, and lets say the car turns a corner fast and pulls around lateral 3g. Since under normal condition car weights 1000 x 9.81 (1g) = 98100 Newtons , will the car now weight 1000 x 9.81 x 3 (3g) = 294300 Newtons while pulling 3g during cornering?
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$\begingroup$ In what direction are you defining weight? Down? Or perpendicular to the road surface? And is the road corner sloped like a race track or is if horizontally? $\endgroup$– DKNguyenCommented Oct 6, 2022 at 16:00
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$\begingroup$ road isnt sloped just normal road without any slope or banking etc. I also meant the downwards weight. $\endgroup$– Timothy SchererCommented Oct 6, 2022 at 16:05
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$\begingroup$ I would be surprised if a car could corner at 3 g. I should think not much above 1 g would be the maximum, possibly not even that much. Rubber wheels rolling on pavement are not going to do much better than a coef of friction of 1. $\endgroup$– BillOnneCommented Oct 6, 2022 at 16:17
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3$\begingroup$ The weight of the car will be the same. It's the distribution of the weight that changes. $\endgroup$– Bob DCommented Oct 6, 2022 at 16:34
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$\begingroup$ The weight won't change because gravity of 1g is still the only force acting downwards on the road surface. The equivalent of 3gs of acceleration from the turning is the centripetal force which acts laterally. $\endgroup$– DKNguyenCommented Oct 6, 2022 at 17:10
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1 Answer
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If you're on a flat road and you define the direction of the weight to be vertically downwards(parallel to $\vec{g}:gravity) $; then no, the weight will not change, it'll be $mg$ throughout. If you turn fast enough on the flat road and define the direction of the weight to be radially outwards then it may become more than $mg$ (centrigual force), but most likely the car will topple before reaching that point as friction may not be able to counter the centrifugal force.
Chances are better on a banked road. If you're on a banked road and driving fast enough, and you define the direction of the weight to be $\perp$ the road, then a component of centrifugal force may make the weight more than the normal weight.(you may define the direction of the weight to be radially outwards too, it can work. Which one of these directions works more easily will depend on angle of banking, coefficient of friction, radius of the turn etc.). The car may topple but not as easily as it would've on a flat road. If even at a banked road you define the direction of the weight to be vertically downwards, the weight will not change, it'll stay $mg$.