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Chapter 12: Problem 8

Does choosing a pivot point in an equilibrium problem mean that something is necessarily going to rotate about that point?

Short Answer

Expert verified
No, choosing a pivot point in an equilibrium problem does not necessarily mean that something is going to rotate about that point. The pivot point is just the point from which torques are calculated.

Step by step solution

01

Understanding Pivot Points

In the context of equilibrium problems in physics, a pivot point is a point of rotation or balance. This is where all forces are considered to act upon.
02

Understanding Principles of Rotational Motion

An object is said to be in rotational equilibrium if the sum of the torques acting on it is equal to zero. In other words, the object won't start to rotate.
03

Applying Knowledge to Address the Question

Just because a pivot point is chosen in an equilibrium problem, it doesn't necessarily mean that something will rotate about that point. It is merely a point about which one calculates the torque. The object under consideration may or may not rotate about the pivot point depending on whether the conditions for equilibrium are met.

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Most popular questions from this chapter

If you take the pivot point at the application point of one force in a static- equilibrium problem, that force doesn't enter the torque equation. Does that make the force irrelevant to the problem? Explain.

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A body is subject to three forces: \(\vec{F}_{1}=1 \hat{\imath}+2 \hat{\jmath} \mathrm{N},\) applied at the point \(x=2 \mathrm{m}, y=0 \mathrm{m} ; \vec{F}_{2}=-2 \hat{\imath}-5 \hat{\jmath} \mathrm{N},\) applied at \(x=-1 \mathrm{m}, y=1 \mathrm{m} ;\) and \(\vec{F}_{3}=1 \hat{\imath}+3 \hat{\jmath} \mathrm{N},\) applied at \(x=-2 \mathrm{m}, y=5 \mathrm{m} .\) Show that (a) the net force and (b) the net torque about the origin are both zero.

Give an example of an object on which the net force is zero, but that isn't in static equilibrium.
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