Problem

In Prob. $16.78,$ determine $(a)$ the distance $h…

Question

Answered step-by-step

Problem 78 Easy Difficulty

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 $A$

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Averell Hause

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Physics 101 Mechanics

Vector Mechanics for Engineers: Statics and Dynamics

Chapter 16

Plane Motion of Rigid Bodies: Forces and Accelerations

Section 2

Constrained Plane Motion

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 16

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Problem 101

Problem 102

Problem 103

Problem 104

Problem 105

Problem 106

Problem 107

Problem 108

Problem 110

Problem 111

Problem 112

Problem 113

Problem 114

Problem 115

Problem 116

Problem 117

Problem 118

Problem 119

Problem 120

Problem 121

Problem 122

Problem 123

Problem 124

Problem 125

Problem 126

Problem 127

Problem 128

Problem 129

Problem 130

Problem 131

Problem 132

Problem 133

Problem 134

Problem 135

Problem 136

Problem 137

Problem 138

Problem 139

Problem 140

Problem 141

Problem 142

Problem 143

Problem 144

Problem 145

Problem 146

Problem 147

Problem 148

Problem 149

Problem 150

Problem 151

Problem 152

Video Transcript

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.

Get More Help with this Textbook

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

In Prob. $16.78,$ determine $(a)$ the distance $h…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Vector Mechanics for Engineers: Statics and Dynamics

Related Topics

Discussion

Top Physics 101 Mechanics Educators

Elyse Gonzalez

Andy Chen

Liev Birman

Aspen Fenzl

Physics 101 Mechanics Courses

Recommended Videos

Watch More Solved Questions in Chapter 16

Video Transcript

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Andy Chen

Liev Birman

Aspen Fenzl

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Andy Chen

Liev Birman

Aspen Fenzl

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek Vector Mechanics for Engineers: Statics and Dynamics

Elyse Gonzalez

Andy Chen

Liev Birman

Aspen Fenzl

Ferdinand P. Beer, E. Russell Johnston, Jr., David F. Mazurek

Vector Mechanics for Engineers: Statics and Dynamics