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In Prob. $16.78,$ determine $(a)$ the distance $h$ for which the horizontal component of the reaction at $A$ is zero, $(b)$ the corresponding angular acceleration of the rod.

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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

University of Washington

Simon Fraser University

McMaster University

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.

03:26

In Prob. 16.78 , determine…

04:07

In Prob. 16.76 , determine…

07:50

The bent rod has a mass of…

03:34

13:06

A slender rod of length $l…

07:54

A long, uniform rod of len…

05:09

So here this is from the previous question. Using the same free body diagrams, we want to find a value for H such that the horizontal component of the reaction force at Point A is equal to zero and so weakens first, say not to the linear for party. The linear acceleration equals the angular acceleration multiplied by the length divided by two. The moment of inertia this would be for a rod equals and l squared, divided by 12. The force of magnitude P equals the mass times acceleration again, given that the or is also component of the reaction force that a equals 00 Newton's. And this would be the application of Newton's second law. And so we can say that then P Equalling and Alfa over to Ensign and in solving for Alfa Alfa would be to pee divided by M O. Then we can actually solve. Alfa would then be equaling 22 multiplied by 1.5 towns divided by the mass of £4 divided by 32.2 feet for a second squared multiplied by three feet, and this is giving us 8.5 radiance her second squared. We can then take this is actually happens to be our answer for part B. So continuing on for part a. Ah, we can then take the some of the moments considering Clockwise is positive about the center of mass or point G. This would be equal to the sum of the effective moments about the center of Mass or about Point G. And so we can say that then p multiplied by H minus. I'll over to equals that in the moment of inertia multiplied by the angular acceleration, we can then say that p multiplied by H minus, I'll over to this is going to be equal to em l squared, divided by 12 multiplied by the angular acceleration in this case to pee divided by and out. And so we find that this is equaling. Then pl over six the peas then cancel out and we are left with H minus. L divided by two equaling L over six h than equals l over to plus al over six equaling than to L over three. And we can then say that if Alice three feet, this would be to times three feet divided by three, and H is gonna equal then two feet or we can say 24 inches. That is the end of the solution. Thank you for watching.

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