Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 23P from Chapter 7 from Hibbeler's Engineering Mechanics.
Problem 23P
Step-by-Step Solution
We are given that the force applied on pulley is $F = 400\;{\rm{N}}$.
We are asked to calculate the internal force, shear force and moment at point C.
We have the horizontal distance between point A and C is ${d_1} = 1.5\;{\rm{m}}$.
We have the horizontal distance between point A and pulley is ${d_2} = 3\;{\rm{m}}$.
We have the distance between pulley and point B is ${d_3} = 2\;{\rm{m}}$.
We have the vertical distance of pulley is $h = 1\;{\rm{m}}$.
We have the diameter of pulley is $d = 0.2\;{\rm{m}}$.
Step 2
The free body diagram of whole beam is shown as:
Step 3
Applying the moment of force equation about point B:
\[ - {A_y}\left( {{d_2} + {d_3}} \right) - F\left( {h + d} \right) = 0\]Step 4
Substitute the known values in the equation:
\[\begin{array}{c} - {A_y}\left( {3\;{\rm{m}} + {\rm{2}}\;{\rm{m}}} \right) - \left( {400\;{\rm{N}}} \right)\left( {1\;{\rm{m}} + {\rm{0}}{\rm{.2}}\;{\rm{m}}} \right) = 0\\ - {A_y}\left( {5\;{\rm{m}}} \right) = \left( {400\;{\rm{N}}} \right)\left( {1.2\;{\rm{m}}} \right)\\ {A_y} = \frac{{ - \left( {400\;{\rm{N}}} \right)\left( {1.2\;{\rm{m}}} \right)}}{{\left( {5\;{\rm{m}}} \right)}}\\ = - 96\;{\rm{N}} \end{array}\]Step 5
Applying the equilibrium force of equation along x-axis:
\[\begin{array}{c} \sum {F_x} = 0\\ {A_x} + F = 0\\ {A_x} = - F \end{array}\]Step 6
Substitute the known values in the equation:
\[{A_x} = - 400\;{\rm{N}}\]Step 7
The free body diagram for point C is shown as:
Step 8
Applying the equilibrium force of equation along x-axis at point C:
\[\begin{array}{c} \sum {F_x} = 0\\ {A_x} + {N_C} = 0\\ {N_C} = - {A_x} \end{array}\]Step 9
Substitute the known values in the equation:
\[\begin{array}{c} {N_C} = - \left( { - 400\;{\rm{N}}} \right)\\ = 400\;{\rm{N}} \end{array}\]Step 10
Applying the equilibrium force of equation along y-axis:
\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} - {V_C} = 0\\ {V_C} = {A_y} \end{array}\]Step 11
Substitute the known values in the equation:
\[{V_C} = - 96\;{\rm{N}}\]Step 12
Applying the moment of force equation about point C:
\[\begin{array}{c} {M_C} - {A_y}{d_1} = 0\\ {M_C} = {A_y}{d_1} \end{array}\]Step 13
Substitute the known values in the equation:
\[\begin{array}{c} {M_C} = \left( { - 96\;{\rm{N}}} \right)\left( {1.5\;{\rm{m}}} \right)\\ = - 144\;{\rm{N}} \end{array}\]