Share
4–161. Replace the loading by an equivalent resultant force and couple moment acting at point O.
ReportQuestion
Please briefly explain why you feel this question should be reported.
4–161. Replace the loading by an equivalent resultant force and couple moment acting at point O.
Answers ( 2 )
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this answer should be reported.
Beam Problem Solution
body {
font-family: Arial, sans-serif;
margin: 20px;
}
h1 {
color: #2c3e50;
}
.step-container {
margin: 10px 0;
position: relative;
}
.step-number {
background-color: red; /* red background for step number */
color: white; /* white text for the step number */
padding: 5px 10px;
border-radius: 5px 5px 0 0; /* round top corners */
position: absolute; /* position it above the box */
left: 0; /* align to the left */
right: 0; /* align to the right */
text-align: center; /* center the text */
}
.step-box {
border: 2px solid red; /* red border */
background-color: white; /* white background */
padding: 15px;
border-radius: 5px; /* slight rounding for aesthetic */
margin-top: 30px; /* space for step number */
}
.equation {
font-size: 1.2em;
margin: 10px 0;
}
ul {
padding-left: 20px;
}
li {
margin-bottom: 5px;
}
Engineering Mechanics Problem Solution
We are asked to obtain the horizontal and vertical components of reaction at point A and the tension in cable BC.
\sum F_x = 0 \quad \Rightarrow \quad A_x – F_{BC} \cos \phi = 0 \quad \Rightarrow \quad A_x = F_{BC} \cos \phi
\]
\sum F_y = 0 \quad \Rightarrow \quad A_y – W + F_{BC} \sin \phi = 0 \quad \Rightarrow \quad A_y = W – F_{BC} \sin \phi
\]
\left\{ W \left( \frac{l}{2} \cos \theta \right) + \left( F_{BC} \cos \phi \right) \left( l \sin \theta \right) – \left( F_{BC} \sin \phi \right) \left( l \cos \theta \right) \right\} = 0
\]
\frac{Wl}{2} \cos \theta – F_{BC} l \left( \sin \phi \cos \theta – \cos \phi \sin \theta \right) = 0
\]
F_{BC} = \frac{-W \cos \theta}{2 \sin(\phi – \theta)}
\]
A_x = \left( \frac{W \cos \theta}{2 \sin(\phi – \theta)} \right) \cos \phi = \frac{W \cos \theta \cos \phi}{2 \sin(\phi – \theta)}
\]
A_y = W – \frac{W \cos \theta}{2 \sin(\phi – \theta)} \sin \phi
\]
= W \left( 1 – \frac{\cos \theta \sin \phi}{2 \sin(\phi – \theta)} \right)
\]
= W \left( \frac{2 \sin \phi \cos \theta – 2 \cos \phi \sin \theta – \cos \theta \sin \phi}{2 \sin(\phi – \theta)} \right)
\]
= W \left( \frac{\sin \phi \cos \theta – 2 \cos \phi \sin \theta}{2 \sin(\phi – \theta)} \right)
\]