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Block Tipping Over [Torque and Forces]

  • Thread starter fobbz
  • Start date
27
0
1. Homework Statement
A rectangular block 3m high and 1.8m wide weighing in at 7600N is pushed with a force 20.5° to the height of the block 1.6m from the bottom of the block.

Calculate the Force Required to tip over this block.


2. Homework Equations

Torque = perpendicular force x distance or force x perpendicular distance
3. The Attempt at a Solution

Firstly, I make a free body diagram of the block.

N force up
mg force down
F 69.5° [N o E]

Secondly, I write my ƩTorque eq, using the pivot point where the box will tip on an angle. I will call this P.

Ʃτp = 0
(0.9)mg = (Fsin69.5)(1.6)
F = 4564.0 N
 

BruceW

Homework Helper
3,609
119
I'm not certain I understand what angle the applied force is meant to be at... is it meant to be 69.5 degrees from the horizontal?

I don't think the equation is quite right. The torque due to gravity is correct, but I think the torque due to the applied force should be different.

The torque is the cross product of position (relative to the pivot) with the force. The position of the applied force will be on the other side of the block compared to the pivot point. And the fact that the force is applied 1.6m from the bottom makes the calculation a bit trickier.

To work out the torque due to the applied force, you could either calculate the distance which crosses the line of the force perpendicularly and times this with the full force F. (This will take a few calculations of angles).

Or you could do the cross product, which I think will be easier.
 

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