What force does the road exert on a car as the car passes the highest point of a bump?

Question

The mass of the car is 1720 kg; the radius of the circe is 42.1 m (imagine that the bump is shaped as part of a circle); the car travels at 16.6 m/s... I've been trying to figure it out, but can't get the right answer. Help needed. Thanks in advance.

That's what I've been thinking, but that's not the right answer. It's not mg + (mv^2 / r) like all you guys have been saying.... this is so frustrating. Any other ideas??

Answer 1

The upward reaction force (at the top of the bump) is NOT mg + mv^2/r.

Per unit mass, it is g - v^2/r. Note the sign.

Also, the relevant mass is not the whole mass of the car m but, say, about *half* of it---the rear wheels are not even in contact with the bump unless it's a really huge one (was this overlooked when the question was set?)

The previous answers are *almost* correct---except for the sign mistake (+ instead of - in the centripetal term). Taking the upward direction as +ve, the gravity force is -ve (-g per unit mass), the bump's reaction n is +ve, and the centripetal force is -ve (c = -v^2/r per unit mass), with c = -g + n (gravity and reaction add up to provide the req'd centripetal acceleration, which is negative, i.e. downward). So, per unit mass, the reaction is

n = -gravity + centripetal = g - v^2/r.

IOW, at that point, the bump's curvature actually lessens the load on the wheels (it's a different story at the point of initial contact of course).

The relevant mass for the front wheels is m/2, so the numerical answer is:

Reaction on the *front* wheels is
(g - v^2/r)m/2 = (9.8 - (16.6)^2/42.1)1720/2
= 2800 (newtons) = 286 kG (kilograms-force)

[An answer that forgets to halve the mass would be 5600 N.]

Reaction on the *rear* wheels is: gm/2 = 9.8*1720/2 = 8428 (newtons) = 860 kG (kilograms-force)


PS: I see now that the correct sign did also occur to gp4rts in his last edit.

Answer 2

You must imagine the car travelling in a vertical circle over the bump. There will be a centrifugal force on the car which comes from the road, since the road is causing the angular acceleration of the car. Centrifugal force is given by M*v^2/R; you have all the numbers for that. In addition, you must add the weight of the car to that value (weight = M*g). Even in the absence of the bump, the road is pushing up on the car that amount, so the centrifugal force adds to that.

EDIT: The only other way I can figure this is to compute the vertical acceleration produced by the bump. The vertical displacement of the car as it goes over the bump is given by r^2 =x^2 + y^2, the equation for a circle. x = v*t, so y =√[ r^2 - v^2*t^2]. the vertical acceleration is the second derivative of y, y''(t). The force of the road would then be M*g = M*y''. I will do more on this so check again later.

EDIT AGAIN: I just worked this out numerically in Mathcad. I get the same numerical result from the second approach: The vertical acceleration total in both cases comes out 1.667g. i don't know what else to do.

MORE: Perhaps the centrifugal force at the top of the bump is in opposition to gravity, so the road force is mg - mv^2/r. One can imagine the car leaving the road just after the top; that means the force from the road is zero.

FINAL: I believe the problem we have had is that we imagined the road pushing up on the car in order to lift it over the bump. That is intuitive, but not correct. The frictional force of the tires is what lifts the car, and that is applied parallel to the road and does not affect the force normal to the road. However, the gravitational force is affected by the angle of the car on the bump and is given by M*g*cosø, where ø is the angle with the vertical axis centered on the bump. The centrifugal force is M*v^2/R always acts radially outward (perpendicular to the road). The (normal) force on the road is then F = M*g*cosø - M*v^2/R. At the top of the bump cosø = 1, so this becomes simply M*g - M*v^2/R. The car will leave the road if that force goes to zero; this happens when cosø = v^2/g*R.

Answer 3

for circular motion, we use the eqn,

F = (mv^2) / r

to find theforce the road exert on a car, we find the force the car exerts on the road by adding the weight of the car and the centripedal force, F = mg + (mv^2) / r. as newtons third law of motion states that,

if body A exerts a force on body B, then body B exerts an equal but opposte force on body A

Answer 4

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Answer 5

mg + mv^2/r
it is easy when it asks u - at the highest spot, coz ur normal car weight (force) is mg,, downward.
At any other point, centripetal F will be same, but mg will have appropriate component - described by 'sine' component, acting at an angle to the normal....
Therefore, ur weight and centripetal is countered by road - as opposite force and centrifugal...