Calculating Torque for Tipping a Block: How Can I Determine the Required Force?

In summary, the problem involves determining the amount of force required to tip a 2D block when a perpendicular force is applied at 1/2 its height with a fixed bottom corner opposite the applied force. The torque of the applied force can be calculated by multiplying the force with the perpendicular distance, which is half the height of the block. The torque is then opposed by the torque from the weight of the block, which can be considered to act on the center of gravity of the block. By setting the two torques equal to each other, the amount of force required to tip the block can be determined. The formula for this is F = Mg(W/H).
  • #1
jocose
12
1
1. The problem statement, all variables, and given/known data
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If I apply a force perpendicular to the side of a 2D block at 1/2 its height with a fixed bottom corner opposite the applied force how much force will be required to tip the block?:
  • Block Mass = M
  • Block Width = W
  • Block Height = H
  • Applied Force = F
  • Distance between force and pivot = r
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Homework Equations

The Attempt at a Solution


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If I understand correctly the torque of the applied force will equal F⋅ r ⋅ cos( ∠A ). The cos( ∠A ) is used because the force is being applied perpendicular to the side of the block, not the axis it's rotating in. This will reduce the torque by cos( ∠A ).

What I don't understand is how to calculate the opposing force to see if the resulting torque of the applied force is enough to overpower it.

Specifically, I am confused by the fact that the force is being applied half way, and not at the top of the block. How should the weight above the applied force be quantified?

(NOTE: I am not an enrolled student nor do I have any formal training, I am trying to teach myself. I am making an effort to read texts but it's become overwhelming and I'm looking for some guidance )
 
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  • #2
RCosA = W

Does that look right to you?
 
  • #3
Mark the centre of mass on the diagram. Assume the weight force acts at that point.
 
  • #4
CWatters said:
RCosA = W

Does that look right to you?
Apologies but I'm unfamiliar. Is W weight? If so no that does not look correct. I suppose I'm not understanding what you're getting at.
 
  • #5
CWatters said:
Mark the center of mass on the diagram. Assume the weight force acts at that point.
I've marked the center of gravity on the diagram. I've assumed its in the center (half the width and half the height). I'm still unclear how to utilize it in this instance.
 
  • #6
Hi jocose! ;)

Torque is the force times the perpendicular distance (the so called lever arm) to the point of rotation.
The lever arm of the horizontal force ##F## is ##r \cdot \sin( ∠A ) =H/2##.
It is opposed by the torque from the weight of the block, which is ##Mg##, which can be considered to act on the center of gravity (CG).
That weight has a lever arm of ##W/2##.
It means that when the block is about to tip, we must have:
$$F\cdot \frac H 2 = Mg \cdot \frac W 2 \quad\Rightarrow\quad F = Mg \cdot \frac WH$$
 
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Likes jocose and Dr.D
  • #7
jocose said:
Apologies but I'm unfamiliar. Is W weight? If so no that does not look correct. I suppose I'm not understanding what you're getting at.
No not weight. W is the width of the block marked on your diagram. I was hoping you would realize that it's not RCosA you need but RSinA that you need to multiply by F to get the torque.
 

FAQ: Calculating Torque for Tipping a Block: How Can I Determine the Required Force?

1. What is meant by "force required to tip a block"?

The force required to tip a block refers to the minimum amount of force needed to cause a stationary block to start rotating or falling over.

2. How is the force required to tip a block calculated?

The force required to tip a block can be calculated by multiplying the weight of the block by its height and the coefficient of static friction between the block and its supporting surface.

3. What factors affect the force required to tip a block?

The force required to tip a block can be affected by the weight and shape of the block, the surface it is resting on, and the coefficient of static friction between the block and the surface.

4. Can the force required to tip a block be greater than the weight of the block?

Yes, the force required to tip a block can be greater than the weight of the block if the coefficient of static friction between the block and its supporting surface is high enough.

5. How can the force required to tip a block be reduced?

The force required to tip a block can be reduced by decreasing the weight of the block, increasing the coefficient of static friction between the block and its supporting surface, or changing the shape of the block to make it more stable.

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