Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information....

# A flagpole

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A 10-kg flagpole is being held up by a lightweight horizontal cable, and is propped against the foot of a wall as shown in the figure. If the cable is only capable of supporting a tension of 70 N, how great can the angle be without breaking the cable?

All three objects in the figure are supposed to be in equilibrium: the pole, the cable, and the wall. Whichever of the three objects we pick to investigate, all the forces and torques on it have to cancel out. It is not particularly helpful to analyze the forces and torques on the wall, since it has forces on it from the ground that are not given and that we don't want to find. We could study the forces and torques on the cable, but that doesn't let us use the given information about the pole. The object we need to analyze is the pole.

The pole has three forces on it, each of which may also result in a torque: (1) the gravitational force, (2) the cable's force, and (3) the wall's force.

We are free to define an axis of rotation at any point we wish, and it is helpful to define it to lie at the bottom end of the pole, since by that definition the wall's force on the pole is applied at r = 0 and thus makes no torque on the pole. This is good, because we don't know what the wall's force on the pole is, and we are not trying to find it.

With this choice of axis, there are two nonzero torques on the pole, a counterclockwise torque from the cable and a clockwise torque from gravity. Choosing to represent counterclockwise torques as positive numbers, and using the equation |τ| = r |F| sin θ, we have

rcable|Fcable| sin θcable - rgrav |Fgrav | sin θgrav = 0 .

A little geometry gives θcable = 90 ° - α and θgrav = α, so

rcable|Fcable| sin(90 ° - α) - rgrav |Fgrav | sinα = 0 .

The gravitational force can be considered as acting at the pole's center of mass, i.e., at its geometrical center, so rcable is twice rgrav , and we can simplify the equation to read

2|Fcable| sin(90 ° - α) - |Fgrav | sinα = 0 .

These are all quantities we were given, except for , which is the angle we want to find. To solve for we need to use the trig identity sin(90 °- x) = cos x,

2|Fcable| cosα - |Fgrav | sinα = 0 ,

which allows us to find

Last Update: 2010-11-11