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Currently, the formula I am using to calculate the rotational torque applied onto an axis is $T = Fr\sin \theta$, where $T$ is the torque (in $\mathrm{N\ m}$), $F$ is the force acting, $r$ is the distance between the point on the object at which the force is acting and the centre of rotation (the pivot point), and $\theta$ is angle between $F$ and $r$. However, this assumes the force is being applied on a single point.

Is there a formula or method to calculate the torque applied when the force is applied uniformly across the entire object? For example, if a gate of length $200\ \mathrm{cm}$ has a force of $200\ \mathrm{N}$ applied uniformly across it, what would be the torque experienced?

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    $\begingroup$ By integration. $\endgroup$ – lucas May 19 '16 at 10:02
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In the case of the gate, F=200N is the distributed force, which is applied from x=a=0m to x=b=2m from the axis. The force applied on an element of length dx is (F/L)dx Newtons where L=b-a. The torque on the gate due to the force on this element is dT = (F/L)xdx. We integrate this from x=a=0m to x=b=2m to get the total torque :
$T = [\frac{F}{L}\frac12x^2] = \frac12\frac{F}{L}(b^2-a^2) = \frac12 \frac{F}{b-a}(b-a)(b+a) = \frac12(b+a)F = 200Nm.$

In this case, because the force is applied uniformly, the torque is the same as when the total force is applied at the 'centre of force'.

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A force applied uniformly to a uniformly dense object (or with a uniform-force-per-unit-mass even if the object is not uniformly dense) is equivalent to the same total force applied at the center of mass of the object. The equivalence means the same total force and the same total torque.

In particular, if the center of mass is chosen as the origin of the coordinates, the total torque exerted by this uniformly acting force is zero because it, $\tau = \int dm\, \vec r \times \vec f$ is proportional to $\int \vec r \,dm$ =0.

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In your example you have the total load applied as

$$ F = \int \limits_0^\ell w\, {\rm d} x = w \ell $$

where $w$ is the linear force density (in Newtons per meter).

To get the torque you just include the position of the force

$$ \tau = \int \limits_0^\ell x\, w\,{\rm d} x = \frac{1}{2} w \ell^2 $$

The rule for distributed loading is to find the total load and apply it on the centroid of the distribution graph. For uniform loading this is the middle location. The results are the same as above $$\tau =(\text{load})(\text{centroid}) = \left( w \ell \right) \left( \frac{\ell}{2} \right)$$

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