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I know that for a mass to be in static equilibrium two things have to be satisfied: $\sum F=0$ and $\sum T=0$, where $T$ represents torque. However, I am not sure how the latter can apply in a situation such as this

diagram

Here, a beam is is static equilibrium with only two forces acting on it (let us only consider the y-axis for simplicity's sake), $F_g$ and $F_t$. Clearly the rule of $\sum F=0$ holds and this can be proven with a simple $F_{net}$ statement. However, if the pivot is placed at the very end on the right hand side $\sum T = F_t \sin(55)\cdot0 - F_g\cdot x$, where $x$ is some arbitrary non-zero distance. How can this be if we need $\sum T = 0$?

Phoenix87
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Jabbath
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  • If the center of mass is not moving appreciatly, then sum of forces must be zero. If the body is not rotating appreciatly then the sum of torques about any point is zero. – John Alexiou Jan 17 '15 at 00:22
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    You are also forgetting that the reaction force at the wall on the beam might also have a direction unequal to normal to the wall. – fibonatic Jan 17 '15 at 01:14
  • There are three forces in this problem - you're forgetting the force from wall on the beam. – Brionius Jan 17 '15 at 01:45

1 Answers1

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There is also a force from the wall acting on the left end of the beam. This force provides a positive $F_{y_{wall}}$ that helps hold the beam up and a positive $F_{x_{wall}}$ that prevents the beam from swinging to the left.

The torque is then: $T = -F_g x - 2x F_{y_{wall}}$. Setting this equation to zero gives us $F_g = 2 F_{y_{_wall}}$. Note that none of the forces depend on x.

Mark H
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Kevin Kostlan
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    It is recommended you use math formatting with $...$ instead of plain text for readability. See http://physics.stackexchange.com/help/notation – John Alexiou Jan 17 '15 at 00:20