# if my car is stuck in a hole....

by Rock_Sniffer
Tags: hole, stuck
 P: 5 I'm just doing a taught experiment, and i want to be able to prove or not that pushing down on a car that is stuck in a hole(increasing down force) uses less work force than pushing it from the back. edit: what i came up with is- ASSUME my car is 10kg and it takes 100N to get out of the hole, but it is only exerting 60N. so a=6m/s^2 and it is required an extra 4m/s^2 to get out. down force is 9.81*10=98.1N how much more do i need to get out of the hole?
 P: 836 The friction would exactly balance the force exerted by the car to get out (60N) so there would be no acceleration till it crosses the 100N mark(limiting friction) Also, I fail to see how pushing the car downwards would help get it out of the hole. Could you please elaborate it?
 P: 5 down force increases friction... according to my regents physics essentials text book. oh i see we need 100n for the friction to be enough to take it out. so what i am saying is that we need 40n pushing from the back, how much do we need to increase down force to take it out? Down Force= 9.81m/s^2*10kg=98.1N
P: 5

## if my car is stuck in a hole....

okay lets simplify the question. the car can only exert 60N so lets say we need 40N to move it. (10kg)

coefficient of friction is F/W = 40/98.1= .40775
 HW Helper P: 6,164 Just my 2 cents. Usually the problem with being in a hole, is lack of traction. The engine usually has enough power, but without traction it can't climb out. If you increase the weight, you do indeed increase the friction. But you also increase the weight that must be pulled out. This tends to cancel out a bit. Pushing from the back means that the weight is not increased, but the friction against the front of the hole is increased, which is exactly where you need the traction. This is more effective. Of course, it's usually easier to just sit on the part of the vehicle that's in the hole...
 P: 5 so what are you saying, applying force to the part thats in the hole is eaiser than pushing from the back? thats what i am trying to show mathematically. EDIT: i was assuming the car can only exert 60N and it needs 100N to get out, so an additional 40N of force from the back or increasing the down force by X amount so that 60N from the car is sufficient. Find X and see if it is less than 40N; hence less force than pushing from the back.
 HW Helper P: 6,164 I'm going for a qualitative approach before trying to calculate anything. If the hole is deep enough (deeper than the axis of the wheel), it's impossible to get out without pushing from the back, or up. If it's not that deep, it may be possible to sit on it to improve traction, but pushing from the back, or up, can also work. It depends entirely on the type of hole, the slope, and the material involved (mud or snow?). Btw, one of the typical problems is that the other wheel starts spinning, due to the differential in the axis. But there are tricks to handle that too.
P: 5
 Quote by I like Serena I'm going for a qualitative approach before trying to calculate anything. If the hole is deep enough (deeper than the axis of the wheel), it's impossible to get out without pushing from the back, or up. If it's not that deep, it may be possible to sit on it to improve traction, but pushing from the back, or up, can also work. It depends entirely on the type of hole, the slope, and the material involved (mud or snow?). Btw, one of the typical problems is that the other wheel starts spinning, due to the differential in the axis. But there are tricks to handle that too.
hmmm i see. i was just trying to show if pushing from the back is less or more efficient than
increasing downforce. i guess it depends.

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