Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 4RP from Chapter 5 from Hibbeler's Engineering Mechanics.
Problem 4RP
Step-by-Step Solution
We are given that a force of magnitude $F = 50\;{\rm{lb}}$ is applied by the hand on the lever.
We are asked to calculate the horizontal and vertical reaction at pin at A and the reaction of the roller at B.
Step 2
Let ${A_x}$ and ${A_y}$ are the horizontal and vertical components of reaction at point A and ${N_B}$ is the reaction at point B.
Draw a free-body diagram of the given system.
Apply equilibrium equation for the moment about point A:
\[\begin{array}{c} \sum {{M_A}} = 0\\ F\cos 30^\circ \times 20\;{\rm{in}}. + F\sin 30^\circ \times 14\;{\rm{in}}. - {N_B} \times 18\;{\rm{in}}. = 0\\ {N_B} \times 18\;{\rm{in}}. = F\cos 30^\circ \times 20\;{\rm{in}}. + F\sin 30^\circ \times 14\;{\rm{in}}.\\ {N_B} = \frac{{F\left( {\cos 30^\circ \times 20\;{\rm{in}}{\rm{.}} + \sin 30^\circ \times 14\;{\rm{in}}{\rm{.}}} \right)}}{{18\;{\rm{in}}.}} \end{array}\]Substitute $50\;{\rm{lb}}$ for $F$ in the above equation:
\[\begin{array}{c} {N_B} = \frac{{\left( {50\;{\rm{lb}}} \right) \times \left( {\cos 30^\circ \times 20\;{\rm{in}}{\rm{.}} + \sin 30^\circ \times 14\;{\rm{in}}{\rm{.}}} \right)}}{{18\;{\rm{in}}.}}\\ = 67.56\;{\rm{lb}} \end{array}\]Step 3
Apply equilibrium equation of forces in the horizontal direction:
\[\begin{array}{c} \sum {{F_x}} = 0\\ {A_x} - F\sin 30^\circ = 0\\ {A_x} = F\sin 30^\circ \end{array}\]Substitute $50\;{\rm{lb}}$ for $F$ in the above equation:
\[\begin{array}{c} {A_x} = \left( {50\;{\rm{lb}}} \right)\sin 30^\circ \\ = 25\;{\rm{lb}} \end{array}\]Apply equilibrium equation of forces in the vertical direction:
\[\begin{array}{c} \sum {{F_y}} = 0\\ {A_y} - F\cos 30^\circ - {N_B} = 0\\ {A_y} = F\cos 30^\circ + {N_B} \end{array}\]Substitute the value of $F$ and ${N_B}$ in the above equation:
\[\begin{array}{c} {A_y} = \left( {50\;{\rm{lb}}} \right)\cos 30^\circ + \left( {67.56\;{\rm{lb}}} \right)\\ = 110.86\;{\rm{lb}} \end{array}\]