Understanding the Physics Behind Moving a Car with a Rope and a Pole

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Attaching a strong rope to a light pole and a car on a downhill incline allows for easier movement when pulling or pushing perpendicularly on the rope. The discussion highlights the importance of tension in the rope and the angle at which force is applied. Mathematically, the force exerted on the car can be analyzed using trigonometric functions, specifically considering the angle θ between the rope and the direct line to the car. The force required to push the car is greater than the force needed when pulling at an angle, as the latter utilizes components of force more effectively. Understanding these physics principles clarifies why pulling is more efficient than pushing in this scenario.
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Homework Statement


You attach a strong rope or steel cable to a light pole and a car parked on a downhill incline. If you have a student lift, pull, or push perpendicularly on the rope in the center it will be easier to move the car than if you were to push it. Why? Prove this mathematically.


Homework Equations





The Attempt at a Solution


I know it has something to do with the tension, but I can't figure out exactly how it works.
 
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Tearsandrille said:
You attach a strong rope or steel cable to a light pole and a car parked on a downhill incline. If you have a student lift, pull, or push perpendicularly on the rope in the center it will be easier to move the car than if you were to push it. Why? Prove this mathematically.

Hi Tearsandrille! :smile:

(do you mean a lamp post? :wink:)

Hint: Suppose the rope has length L, and is at an angle θ to the direct line between the car and the lamp post … if you push it with force F in the middle, what is the force on the car?
 
I did mean lamp post. You'll have to excuse me there, my brain was friend trying to draw all the FBD and vectors.

So, wouldn't the force on the car be tan(theta) * Fapp.

My trig isn't so great, but how is that less than the force you would need to push the car? I believe in order to push the car it would be m*g*sin(theta).
 
Hi Tearsandrille! :wink:

(just got up :zzz: …)
Tearsandrille said:
So, wouldn't the force on the car be tan(theta) * Fapp.

Nope.

Take components along the line of the Fapp. :smile:
 
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