Height of the force at which an object will tip

In summary, a bale with dimensions 0.25 m x 0.50 m x 0.80 m and mass 30.0 kg is being dragged with constant speed on a horizontal surface by a force \vec{F}. The coefficient of kinetic friction is 0.35, and the magnitude of \vec{F} required for constant speed is 103 N. To prevent the bale from tipping, the torque about the center of rotation must be greater than the torque caused by the weight of the bale. This results in a condition where the leading bottom edge of the bale is forward and the center of mass is behind the pivot point. The value of h at which the bale just
  • #1
vic226
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Homework Statement


A bale is 0.25 m wide, 0.50 m high, and 0.80 m long, with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. It is dragged along a horizontal surface with constant speed by a force [tex]\vec{F}[/tex]. The coefficient of kinetic friction is 0.35.

YF-11-58.jpg


A) Find the magnitude of the force [tex]\vec{F}[/tex].

B) Find the value of h at which the bale just begins to tip.

Homework Equations



[tex]\sum{\vec{F}} = 0[/tex]
[tex]\sum{\vec{\tau}} = 0[/tex]

The Attempt at a Solution



Part A is easy; for the bale to move at constant speed,

[tex]F = fr = \mu F_N = \mu mg = 102.9 N \approx 100 N[/tex]

Part B is where I get stuck. I just randomly submitted the answer based on the picture's scale (I know it's not reliable but I had no choice, as it was a few minutes before due), and I got the answer h = 0.36 m. I know it has something to do with torque at equilibrium and the center of gravity of the system, but I fail to understand how it would work on such situation. Any pointers would be much appreciated.
 
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  • #2
First of all the answer to part A is 103 N. to three significant figures.

The condition for falling is the torque about the centre of rotation - the leading bottom edge of the bale is forward. The centre of mass is behind the pivot point so there is a net backward torque. You have to find the point r such that

[tex]\vec F \times \vec r > m\vec g \times \vec r_{cm}[/tex]

AM
 

1. What is the definition of "Height of the force at which an object will tip"?

The height of the force at which an object will tip, also known as the tipping point, is the point at which an object is no longer stable and will begin to fall or tip over due to an external force acting upon it.

2. How is the height of the force at which an object will tip determined?

The height of the force at which an object will tip is determined by the center of gravity of the object and the position of the external force. The higher the center of gravity and the farther the external force is from the base of the object, the lower the tipping point will be.

3. What factors affect the height of the force at which an object will tip?

The height of the force at which an object will tip is affected by the weight and shape of the object, the surface on which it is placed, and the strength and direction of the external force. The height of the object's center of gravity and the horizontal distance of the external force from the base of the object also play a role.

4. Can the height of the force at which an object will tip be changed?

Yes, the height of the force at which an object will tip can be changed by altering the factors that affect it. For example, lowering the object's center of gravity or moving the external force closer to the base of the object can increase the tipping point.

5. Why is understanding the height of the force at which an object will tip important?

Understanding the height of the force at which an object will tip is important for ensuring safety and stability when working with objects that may tip over. It can also help in making design decisions for structures or objects to prevent tipping and maintain balance.

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