if i have a car jack and put under the car(1000kg) near whichever tire. Is it right to assume that the overall force exerted on the car jack is equal to 1/4 of the weight of the car(2500N)?
2007-05-08 11:51:46 · 2 answers · asked by eddy1234 1 in Science & Mathematics ➔ Physics
Like the other answer said, you'd be right if and only if the weight of the car was evenly distributed over the area of the car. Look at it this way, the weight of the car (W) has to be offset by the four forces touching the floor (three tires and one jack). That is W = sum of the four forces = sum(f). Why? Because if W > sum(f), the car would accelerate downward towards the ground. If W < sum(f), the car would begin to rise and accelerate upward.
When the center of mass C is equidistant from the four points of contact on the ground, the forces will be equal to each other. But if that center of mass shifts to one direction or the other, the contact point closest to the new center will have increased in force, while the contact farthest from that new center will have lost force. This is necessary to keep sum(f) = W.
2007-05-08 12:30:00 · answer #1 · answered by oldprof 7 · 0⤊ 0⤋
Assuming that the car is uniform, then yes because the weight would be evenly distributed over the 4 tires,
if the car is not of uniform weight, you would need to find the center of mass and solve from there.
2007-05-08 12:00:59 · answer #2 · answered by kid. 2 · 0⤊ 0⤋