Another Action/Reaction question Donkey and a cart

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The discussion centers on understanding action-reaction pairs in the context of a donkey pulling a cart. It highlights that the donkey can move the cart if it exerts a greater force than the cart exerts back on it, emphasizing the role of friction with the ground. The forces acting on the donkey and cart must be analyzed in free-body diagrams, focusing on the interaction between the donkey and the ground rather than just the cart. Gravity is acknowledged as a factor, but it is deemed less relevant to the movement in this scenario. Overall, the key takeaway is that friction is essential for the donkey to pull the cart effectively.
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Another Action/Reaction question...Donkey and a cart!

Hey, i am doing this for prety much basic physics but i need to:

Draw all the action-reaction pairs.
If the cart pulls on the mule with the same force that the mule pulls on the cart, how can the mule move?

slide0023_image047.jpg


Can you help me out??

Thanks
John

The mule would pull it if it exerted more force than the cart puts onto the mule, Right?

Also, are there only F1 and -F1, or are there gravity forces that act as action reaction forces?
 
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Objects don't move if the net force on them is zero.

Do the forces in the diagram act on the same object?

Draw the net forces on the objects in the diagram. Think about which way each object would move.

Also, are there only F1 and -F1, or are there gravity forces that act as action reaction forces?
In this case, we can assume that there is friction, because otherwise the mule would get nowhere. It would tread along the ground, but it would not move forward. Gravity is also there, but it is pretty much irrelevant.

The ground pushes the mule forward when the mule pushes against the ground to take steps, because of friction.

This problem is very important to understand, because otherwise free-body diagrams become confusing.
 
so in order for the mule to pull the cart, there would obviously have to be friction, and it would have to have a force greater than the cart.

But what would the diagram look like for this situation?
 
The forces on the horse and the cart that are labeled in the diagram do not affect the horse's moving forward. It is the forces between the horse and the ground that matter.

Think about that when you draw the diagram.
 
well its not the only action and reaction force on the cart
there is another action and reaction force couple
which the donkey is applying
he is applying force downward and backward
and the reaction force is acting forward and upward
so that's why it moves ...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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