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Two identical uniform rods are connected by a pin at $B$ and are held in a horizontal position by three wires as shown. If the wires attached at $A$ and $B$ are cut simultaneously, determine at that instant the acceleration of $(a)$ point $A,(b)$ point $B$
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Physics 101 Mechanics
Plane Motion of Rigid Bodies: Forces and Accelerations
Constrained Plane Motion
Motion Along a Straight Line
Rutgers, The State University of New Jersey
University of Winnipeg
In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.
In mathematics, a position is a point in space. The concept is abstracted from physical space, in which a position is a location given by the coordinates of a point. In physics, the term is used to describe a family of quantities which describe the configuration of a physical system in a given state. The term is also used to describe the set of possible configurations of a system.
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So here we have a free body diagram for the lifted. Both Freebo diagrams for the two beams connected by the cord. We want to find the angular acceleration and of the entire system or in the assembly. And so we can say that first, given these two free bided high grams, we've accounted for the angles, and we can say that 1st 1 taking clockwise to be positive. He some of moments about a within the equal to £4 multiplied by point five feet plus £4 multiplied by 1.25 feet, minus £6 £6 feet or better foot pounds. My apologies, and this will be then be equal to to multiplied by 1 12 again accounting for the moment of inertia for the rod multiplied by £4 divided by 32.2 feet per second square. That would be for the mass multiplied by one feet one foot squared, rather multiplied by the angular acceleration plus £4 divided by 32.2 foot, her seconds feet per second squared, multiplied by 0.5 foot for half a foot squared, multiplied by the angular acceleration plus again £4 divided by 32.2 feet per second squared multiplied by 1.25 feet. Quantity squared Alpha Tau US £4 divided by 32.2 feet. Her second squared multiplied, then by half a foot multiplied by 0.866 quantity squared Alfa. And so we can get that. Then seven minus six uh, seven minus six foot pounds will then be equaling two point 26 915 after be simple file that multiplied by Alfa and so solving. For Alfa, the angular acceleration is unfound to be 3.7 15 Parade Ian's per second squared, so I'd be, for part a report be We have these two free bite of diagrams here, and at this point, we can actually solve for the moment about be instead of the moment about a and self attention man. And so we find that then again, choosing clockwise to be positive with some of the moments about be equals £4 multiplied by 0.25 feet, minus the tension multiplied by one foot multiplied by sine of 30 degrees. This is also going to be equal, then to the effective moment. Let's account for the moment of inertia multiplied by one foot corner B squared Alfa plus again £4 divided by 32.2 feet per second squared multiplied by 1.25 feet, multiplied by point 25 feet. Try to get this a little bit smaller. 0.25 see Alfa Plus then £4 again multiplied, divided by 32.2 feet per second squared. This will be multiplied by half a foot Well played by co sign of 30 degrees multiplied by a point for 33 feet Alfa. And so we find that then this is gonna equal 2.0 seven 246 Alfa Equalling, of course, 0.7267246 multiplied by Alfa that we found in part a So 3.7154 around at the very end. And this is going to then also equal one minus T multiplied by 1/2 0.7246 multiplied by 3.7154 equals point 26922 and solving for tea. Then T is gonna equal one point 462 pounds. So this would be the tension in the court. That is the end of the solution. Thank you for watching
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