Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 100P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 100P
Step-by-Step Solution
We are given that the compression in spring is $s = 20\;{\rm{mm}}$ and the stiffness constant of the spring is $k = 300\;{\rm{N/m}}$.
We are asked to calculate the normal force acting on lobe at C.
We have the distance between G and F is ${d_1} = 40\;{\rm{mm}}$.
We have the distance between F and E is ${d_2} = 25\;{\rm{mm}}$.
Step 2
The free body diagram of system at point G is shown as:
Step 3
To calculate the spring force acting on point G we use the formula:
\[{F_S} = ks\]Step 4
Substitute the known values in the formula:
\[\begin{array}{c} {F_S} = \left( {300\;{\rm{N/m}}} \right)\left( {20\;{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = 6\;{\rm{N}} \end{array}\]Step 5
Applying the equilibrium force of equation along y-axis:
\[\begin{array}{c} \sum {F_y} = 0\\ - {F_G} + {F_S} = 0\\ {F_G} = {F_S} \end{array}\]Step 6
Substitute the known values in the equation:
\[{F_G} = 6\;{\rm{N}}\]Step 7
The free body diagram of the lobe is shown as:
Step 8
Applying the moment of force equation about point F:
\[ - {F_G}{d_1} + T{d_2} = 0\]Step 9
Substitute the known values in the equation:
\[\begin{array}{c} - \left( {6\;{\rm{N}}} \right)\left( {40\;{\rm{mm}}} \right) + T\left( {25\;{\rm{mm}}} \right) = 0\\ T\left( {25\;{\rm{mm}}} \right) = \left( {6\;{\rm{N}}} \right)\left( {40\;{\rm{mm}}} \right)\\ T = \frac{{\left( {6\;{\rm{N}}} \right)\left( {40\;{\rm{mm}}} \right)}}{{\left( {25\;{\rm{mm}}} \right)}}\\ = 9.6\;{\rm{N}} \end{array}\]