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A uniform slender rod of length $L=36$ in. and weight $W=4$ lb hangs freely from a hinge at $A$. If a force $\mathbf{P}$ of magnitude 1.5 lb is applied at $B$ horizontally to the left $(h=L),$ determine $(a)$ the angular acceleration of the rod, $(b)$ the components of the reaction at $A$

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Physics 101 Mechanics

Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

Section 2

Constrained Plane Motion

Motion Along a Straight Line

Cornell University

University of Michigan - Ann Arbor

University of Sheffield

Lectures

04:34

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.

07:57

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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A uniform slender rod of l…

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A long uniform rod of leng…

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The 4 -lb uniform slender …

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A long, uniform rod of len…

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A slender rod of length $l…

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here the two simple rebounded diagrams we have be the linear acceleration here, equaling 1/2 of the anger like celebration, the moment of inertia than equaling then and I'll squared over 12 for a rod here. We're going to say that clockwise this positive and we can say that the sum of the moments about point A equals the sum of the effective moments about point A. We can say that then the force of magnitude p multiplied by the full length the displacement between the point at which the force is applied and the axis of rotation is equal toe. This would be equal then to the mass times a linear acceleration multiplied by L over two for the center of mass. Plus the moment of inertia for the rod itself multiplied by the angular acceleration. And so this would then be equaling two m al Alfa over to multiplied by our over to plus M l Squared Alfa over 12. And so we can then say that p l equals M l squared Alfa over three. And so for party, we can say that the angular acceleration equals three p over ml and this would be equal to three multiplied by £1.5. This would be divided by £4 divided by 32.2 feet, her second squared, and this would be multiplied by three feet. And so we have then that this would be equaling 2 12 0.8 radiance, her second squared clockwise. It's positive. And then for part B, we confined the reaction, the components of the reaction force a day. And so for Part B, we can say going up the sum of forces in the Y direction equals the some of the forces of the effective forces in the Y direction. We can say the reaction force, the vertical reaction force for uh, point a minus w the weight equaling zero. Therefore the reaction force for the vertical component of the reaction forces equaling the weights, equaling £4 for applying Newton's second law in the and in the X direction or horizontally, we have the horizontal component of the reaction force minus force with magnitude p equaling negative em okay. And so we can say that then the X component within be equal to P minus M multiplied by al Alfa over to this is equal to P minus m. I'll over two multiplied by three p over and out. And so we find that then the horizontal, the horizontal component of the reaction force is equal. And then P minus amel over two multiplied by three p over ml. This is equaling negative. P over to this is gonna be willing to negative £1.5 divided by two. And so we find that then the X component rather the horizontal component for the reaction force is Equalling £0.750. We can say to to the left, that is the end of the solution. Thank you for watching.

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