This week’s set is based on the idea of equilibrium, specifically translational and rotational equilibrium. I’ve **uploaded a problem set here**.

An object is in translational equilibrium when no net force acts on it. No net force means the object is not accelerating, and is therefore either at rest or experiencing a constant velocity.

A body in translational equilibrium may have forces acting on it, but the vector sum of those forces must be zero. This may be expressed as:

∑ **F** = 0

Where the symbol sigma indicates “sum of” and the bold F refers to the forces (vectors, which is why it’s in bold) that act upon the body.

When working on these problems it is easiest to establish a set of coordinate axis and to resolve the forces acting on the object into their components acting along those axes. In this way, the vector equation above is replaced by three scalar equations:

∑ F_{x} = 0 ∑ F_{y} = 0 ∑ F_{z} = 0

Of course, a proper choice of directions for the axes will simplify the calculations.

If the lines of forces acting on an object in translational equilibrium intersect at a common point the object will not rotate. These forces are described as concurrent. If the forces are nonconcurrent, the body will spin even though the resultant force is zero.

Torque, represented by the Greek letter tau ( τ )is the measure of the effectiveness of a nonconcurrent force in turning an object about a pivot point. It is directly related to the resultant force and the perpendicular distance between the line of action of the force and the pivot point (called the moment arm, L). Thus

τ= FL

Torque = Force x moment arm

An object is at rotational equilibrium if the net torque acting on it is zero. Similarly to translational equilibrium, we have for rotational equilibrium the equation:

∑ **τ** = 0

Any point may be chosen as the pivot point for calculating torques. If the sum of the torques on an object in translational equilibrium is zero about some pint, it will be zero at any other point in the object.

And that’s it … the problem set will help demonstrate these ideas.

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