Rotation - Minimum force so that the block topple

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The discussion focuses on the mechanics of a block tipping over under applied force. Participants clarify that the normal force does not act through the center of mass, which is crucial for calculating torque. They suggest using the edge farthest from the applied force as the axis of rotation to analyze the torque generated by both the applied force and gravity. The net torque must equal zero for the block to tip, emphasizing the importance of understanding the forces at play. Overall, the conversation highlights the complexities of torque and equilibrium in rotational motion.
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Homework Statement


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The Attempt at a Solution



I have worked my brain out but can't work out the problem

the normal force will act through th center of mass right?

if not please some one tell me why !
 

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Consider the edge far from the applied force (i.e. the one against the table that it would intuitively tip around) as an axis of rotation...then you can find the torque applied by the force at that point a distance from the axis at an angle (associated with the geometry of the cube). Also, remember gravity would provide another torque at the center of mass, which is also another distance from the axis at a different angle. Hope that helps some.
 
See, there must be a force acting at the bottom which does not allow the cube to slide. However, you do not know its magnitude. So the best option is to calculate torque about the point which does not move. What will be the forces acting? Calculate their torque. Net torque should be just zero such that it tips.

The normal force will not act through the center of the mass.
If it did then- imagine- net torque would be zero and any small amount (even air flow) will tip the cube.:rolleyes:
 
If you could help me solve this problem that'll help me ...

attachment.php?attachmentid=33639&stc=1&d=1301304073.jpg


I can't figure out this problem
 

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okay, let us consider that the normal force is acting through the center of mass. Then the given force acts. As I said in my previous post any amount of force should be enough to tilt the block.
So, the block will tilt(just a bit) and the block will lose contact with ground (except the rightmost edge). Where will the normal force act then?
 
OH yes ... you are right ...

Thanks guys for help ...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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