![]() ![]() At all times, the static equilibrium conditions given by Equation 12.7 through Equation 12.9 are satisfied. View this demonstration to see two forces act on a rigid square in two dimensions. Thus, for planar problems with the axis of rotation perpendicular to the xy-plane, we have the following three equilibrium conditions for forces and torques: With this choice of axis, the net torque has only a z-component, all forces that have non-zero torques lie in the xy-plane, and therefore contributions to the net torque come from only the x- and y-components of external forces. The standard procedure is to adopt a frame of reference where the z-axis is the axis of rotation. For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. In the most general case, equilibrium conditions are expressed by the six scalar equations ( Equation 12.3 and Equation 12.6). The origin of a selected frame of reference is called the pivot point. In one frame of reference, the mathematical form of the equilibrium conditions may be quite complicated, whereas in another frame, the same conditions may have a simpler mathematical form that is easy to solve. Our choice of reference frame is dictated by the physical specifics of the problem we are solving. The practical implication of this is that when applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame. ![]() Hence, we see that the net torque in any inertial frame of reference S ′ S ′ is zero, provided that both conditions for equilibrium hold in an inertial frame of reference S. In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S, the first term vanishes because of Equation 12.5 and the second term vanishes because of Equation 12.2. ∑ k τ → ′ k = ∑ k r → ′ k × F → k = ∑ k ( r → k − R → ) × F → k = ∑ k r → k × F → k − ∑ k R → × F → k = ∑ k τ → k − R → × ∑ k F → k = 0 →. ![]() From our study of relative motion, we know that in the new frame of reference S ′, S ′, the position vector r → ′ k r → ′ k of the point where the force F → k F → k is applied is related to r → k r → k via the equation ![]() Suppose vector R → R → is the position of the origin of a new inertial frame of reference S ′ S ′ in the old inertial frame of reference S. The explanation for this is fairly straightforward. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. Therefore, torque depends on the location of the axis in the reference frame. However, the second condition involves torque, which is defined as a cross product, τ → k = r → k × F → k, τ → k = r → k × F → k, where the position vector r → k r → k with respect to the axis of rotation of the point where the force is applied enters the equation. The first condition involves only forces and is therefore independent of the origin of the reference frame. The first and second equilibrium conditions are stated in a particular reference frame. The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.Īccording to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or Because the motion is relative, what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Notice that the distinction between the state of rest and a state of uniform motion is artificial-that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. Explain how the conditions for equilibrium allow us to solve statics problems.Draw a free-body diagram for a rigid body acted on by forces.Identify the physical conditions of static equilibrium.By the end of this section, you will be able to: ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |