Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 57P from Chapter 3 from Hibbeler's Engineering Mechanics.
Problem 57P
Step-by-Step Solution
We are given the maximum tension ${T_m} = 50\;{\rm{N}}$.
We are asked to determine the greatest weight of the flower pot.
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,
\[\begin{array}{c} \Sigma {F_y} = 0\\ - {F_{AD}}\cos 30^\circ \cos 30^\circ - {F_{AC}}\cos 30^\circ \cos 30^\circ + {F_{AB}}\cos 45^\circ = 0\\ {F_{AC}} = 0.47{F_{AB}} \end{array}\] ....... (2)Here, we can see that the force on AB is greatest and it can be considered as the maximum tension force that is ${T_m} = {F_{AB}} = 50\;{\rm{N}}$.
Step 3
To find the magnitude of force in z direction we will use the relation of equilibrium,
\[\begin{array}{c} \Sigma {F_z} = 0\\ {F_{AD}}\cos 30^\circ \sin 30^\circ + {F_{AC}}\cos 30^\circ \sin 30^\circ + {F_{AB}}\sin 45^\circ - W = 0\\ 0.43{F_{AD}} + 0.43{F_{AC}} + 0.707{F_{AB}} = W \end{array}\] ....... (3)On plugging the values from equation (2) and (1) in equation (3), we get,
\[\begin{array}{c} 0.43{F_{AC}} + 0.43{F_{AC}} + 0.707{F_{AB}} = W\\ 0.43\left( {0.47{F_{AB}}} \right) + 0.43\left( {0.47{F_{AB}}} \right) + 0.707{F_{AB}} = W\\ 0.43\left( {0.47} \right)\left( {50\;{\rm{N}}} \right) + 0.43\left( {0.47} \right)\left( {50\;{\rm{N}}} \right) + 0.707\left( {50\;{\rm{N}}} \right) = W\\ W = 55.56\;{\rm{N}} \end{array}\]