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So imagine a car that weighs 2 tons and thus exerts 20K Newtons on the ground is standing on soft mud and starts sinking into the ground, in this situation, 'the car sinks because the mud can't provide the same 20K Newton Force upwards on the car like how a concrete ground would due to an imbalance of forces (and thus starts accelerating downwards)', thats what I found about this problem online.

But by Newtons third law, the mud should be able to exert an equal and opposite force on the car, yet it dosen't, why is this so?

Qmechanic
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Feraminecarts
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3 Answers3

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But by Newtons third law, the mud should be able to exert an equal and opposite force on the car, yet it dosen't, why is this so?

You are confused by action-reaction pairs. To better understand this, draw a complete free-body diagram that includes car, Earth, and mud.

There are two forces acting on the car in your example:

  • gravitational force from Earth to car; reaction force is gravitational force from car to Earth
  • normal (upward) force from mud to car; reaction force is normal (downward) force from car to mud

Do gravitational and normal force make action-reaction pair? No, they do not. The resultant force on the car is the difference between the two.

Marko Gulin
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  • but isn't the normal force just the reaction of the weight of the car? because in my school we were taught that N is always equal to either mg or mg cosθ, so that would mean that N = mg of the car. please correct what is wrong in my understanding – Feraminecarts Jul 28 '22 at 10:25
  • @Feraminecarts N=mg is a special case of vertical equilibrium. (N=mgcosθ is again, for EQUILIBRIUM on an incline along the perpendicular to the incline.) Consider yourself in a lift accelerating upwards with acceleration a. Then N-mg=ma. So they are not equal. Also, action reaction forces act on different bodies. Here, weight of the body (earth pulling body towards its center) and normal force on body (by ground on body) are acting on the same body and hence canNOT be action-reaction pairs. – insipidintegrator Jul 28 '22 at 11:05
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. . . . . the mud should be able to exert an equal and opposite force on the car, yet it doesn't, . . . . .

Oh, yes it does!

The weight of the car is more than the upward force on the car due to the mud and that is why the car accelerates downwards

The car exerts an equal downward force on the mud (N3L) which causes the mud to compress.

In doing so one could imagine that upward force due to the mud increasing (cf spring compressing} thus reducing the acceleration of the car until it is sufficient to stop the motion of the car.

At all time the force on the car due to the mud is equal in magnitude and opposite in direction to the force on the mud due to the car.

Farcher
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You need to be very careful when applying Newton's third law. When you say that the car exerts a force of 20kN on the ground, what that means is that there is a force of attraction of 20KN between the car as a whole and the Earth as a whole. If you are not considering the interactions at that level, you need to think about the underlying detail.

Specifically, if the car is placed on soft mud, the force the car exerts on the mud is not initially 20KN. As gravity pulls the car downwards, the motion of the car imparts an increasing force on the mud, causing the mud to compress. The force exerted by the car is just the force required to compress the mud. That force increases as the mud gets more and more compressed, until eventually it reaches the point at which it equals 20KN, and then, and only then, the car stops sinking and reaches an equilibrium point at which the attraction due to gravity is balanced by a 20KN reaction from the mud. Up until that point is reached, the force of the car on the mud is less than 20KN.

Marco Ocram
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