What is the critical angle for a box to topple?

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In summary, when the center of mass of the cuboid is over the pivot point, there is no torque exerted in either direction around the pivot.
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songoku
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Homework Statement


As shown in the figure below, a cuboid with uniform density is tilted by hand so that one edge remains in contact with a roughly surfaced floor. The angle of tilt, θ , is gradually increased, and when becomes larger than θ0 , the cuboid leaves the fingers and rotates. As indicated in the figure, the lengths of the two edges are a and b. The edge in contact with the floor does not slide. What is tan θ0?
Untitled%201_zpsnogepyor.png

Homework Equations


∑ torque = 0
torque = F . perpendicular distance

The Attempt at a Solution


I am stuck from the beginning. I try to draw the free body diagram. There are three forces acting on the box; weight, normal force and upward force from hand.

Taking the vertex of the box where it touches the floor as pivot, the normal force won't produce torque and when the cuboids leaves the fingers and rotates, the upward force from the hand also zero. Maybe weight also won't produce torque because when the box rotates, the line of force of weight passes through the pivot. I also don't know the value of the upward force from the hand.

How to set up the equation?
 
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  • #2
songoku said:
Maybe weight also won't produce torque because when the box rotates, the line of force of weight passes through the pivot.
That's the key point: When the box's center of mass is directly over the pivot the box exerts no torque in either direction around the pivot point. The box is then in an unstable equilibrium and could fall either way... :wink:
 
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  • #3
gneill said:
That's the key point: When the box's center of mass is directly over the pivot the box exerts no torque in either direction around the pivot point. The box is then in an unstable equilibrium and could fall either way... :wink:

I get it. We don't need to use equation of torque, just simple trigonometry. The answer is a/b

Thanks a lot
 

1. What is the angle at which a box will topple?

The angle at which a box will topple depends on several factors, including the weight and shape of the box, the surface it is resting on, and any external forces acting on it. In general, a box will topple when its center of gravity shifts outside of its base of support.

2. How can I calculate the angle to topple a box?

To calculate the angle at which a box will topple, you will need to know the weight and dimensions of the box, as well as the coefficient of friction between the box and the surface it is resting on. Using these variables, you can use the equations for torque and center of mass to determine the angle at which the box will topple.

3. Can the angle to topple a box be modified?

Yes, the angle to topple a box can be modified by changing the weight or shape of the box, altering the surface it is resting on, or applying external forces to the box. For example, a heavier box will require a larger angle to topple, while a box with a wider base will be more stable and have a larger angle to topple.

4. What happens if the angle to topple a box is exceeded?

If the angle to topple a box is exceeded, the box will lose its balance and fall. The direction and speed of the fall will depend on the angle and direction of the external force that caused the box to topple.

5. Are there any real-life applications for understanding the angle to topple a box?

Understanding the angle to topple a box is important in many fields, including engineering, construction, and product design. It can help prevent accidents and ensure the stability and safety of structures and objects. For example, engineers use this concept when designing buildings and bridges to ensure they can withstand external forces without toppling over.

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