Calculating Maximum Torque on Rotating Square

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To calculate the maximum torque on a rotating square with a side length of 0.69 m and a force of 17.1 N applied in its plane, the correct approach involves using the formula for torque, which is the force multiplied by the distance from the axis of rotation. The maximum distance from the center to the edge of the square is the diagonal divided by 2, calculated as (0.69 m / √2). The torque can be expressed as 17.1 N multiplied by this maximum distance. Clarification is needed on the correct application of the formula to achieve the desired result.
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A square, 0.69 m on a side, is mounted so that it can rotate about an axis that passes through the center of the square. This axis is perpendicular to the plane of the square. A force of 17.1 N lies in this plane and is applied to the square. What is the magnitude of the maximum torque (in N*m) such that a force could produce?

I have no idea how to approach this problem.

I tried doing Force x max distance from axis
so..
17.1 N x \sqrt{}2[/tex(0.69 / 2)<br /> <br /> but i was not getting the correct answer... any ideas?
 
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hi physics_10! :smile:

(have a square-root: √ :wink:)
physics_10 said:
A square, 0.69 m on a side, is mounted so that it can rotate about an axis that passes through the center of the square. This axis is perpendicular to the plane of the square. A force of 17.1 N lies in this plane and is applied to the square. What is the magnitude of the maximum torque (in N*m) such that a force could produce?

I have no idea how to approach this problem.

I tried doing Force x max distance from axis
so..
17.1 N x \sqrt{}2[/tex(0.69 / 2)<br /> <br /> but i was not getting the correct answer... any ideas?
<br /> <br /> (ie 17.1 x 0.69/√2 ?)<br /> <br /> looks right to me <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" />
 
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