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

[jocose]

1. The problem statement, all variables, and given/known data

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

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 )

 

[CWatters]

RCosA = W

Does that look right to you?

 

[CWatters]

Mark the centre of mass on the diagram. Assume the weight force acts at that point.

 

[jocose]

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.

 

[jocose]

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.

 

[I like Serena]

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:
$$F\cdot \frac H 2 = Mg \cdot \frac W 2 \quad\Rightarrow\quad F = Mg \cdot \frac WH$$

 

[CWatters]

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.