Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 56P from Chapter 3 from Hibbeler's Engineering Mechanics.
Problem 56P
Chapter:
Problem:
The 25-kg flowerpot is supported at A by the three cords. Determine...
Step-by-Step Solution
Step 1
Step 2
Step 3
We are given the mass of flower pot is $m = 25\;{\rm{kg}}$.
We are asked to determine the force acting in each cord.
Step 2
The free body diagram of picture is shown below:
To find the magnitude of forces in x direction we will use the relation of equilibrium,
\[\begin{array}{c} \Sigma {F_x} = 0\\ {F_{AD}}\cos 30^\circ \sin 30^\circ - {F_{AC}}\cos 30^\circ \sin 30^\circ = 0\\ {F_{AD}} = {F_{AC}} \end{array}\] ....... (1)To find the magnitude of force in y direction we will use the relation of equilibrium,
\[\begin{array}{c} \Sigma {F_y} = 0\\ 2{F_{AC}}\cos 30^\circ \sin 60^\circ - {F_{AB}}\sin 45^\circ = 0\\ 1.5{F_{AC}} = 0.707{F_{AB}}\\ {F_{AB}} = 2.12{F_{AC}} \end{array}\] ....... (2)Step 3
To find the magnitude of force in z direction we will use the relation of equilibrium,
\[\begin{array}{c} \Sigma {F_z} = 0\\ 2{F_{AC}}\cos 30^\circ \cos 60^\circ + {F_{AB}}\cos 45^\circ - W = 0\\ 2{F_{AC}}\cos 30^\circ \cos 60^\circ + {F_{AB}}\cos 45^\circ - \left( {25\,{\rm{kg}}} \right)\left( {9.81} \right) = 0\\ 0.866{F_{AC}} + 0.707{F_{AB}} = 245.25\;{\rm{N}} \end{array}\] ....... (3)On plugging the values from equation (2) to equation (3), we get,
\[\begin{array}{c} 0.866{F_{AC}} + 0.707\left( {2.12{F_{AC}}} \right) = 245.25\;{\rm{N}}\\ {F_{AC}} = 103.9\;{\rm{N}} \end{array}\]To find the magnitude of force on AB we will use the relation,
\[\begin{array}{c} {F_{AB}} = 2.12\left( {103.9\;{\rm{N}}} \right)\\ {F_{AB}} = 220.2\,{\rm{N}} \end{array}\]The tension in AD is,
\[\begin{array}{l} {F_{AD}} = {F_{AC}}\\ {F_{AD}} = 103.9\;{\rm{N}} \end{array}\]