Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 90P from Chapter 6 from Hibbeler's Engineering Mechanics.
Problem 90P
Chapter:
Problem:
The pipe cutter is clamped around the pipe P. If the wheel at A...
Step-by-Step Solution
Step 1
Step 2
We are given the force ${F_A} = 80\;{\rm{N}}$ and radius of the wheel and pipe is $r = 7\;{\rm{mm}}$ and $r' = 10\;{\rm{mm}}$ respectively.
We are asked to determine the normal forces of wheel B and C.
Step 2
The angle made by the force is calculated as:
\[\begin{array}{c} \sin \theta = \left( {\frac{{r'}}{{r' + r}}} \right)\\ \sin \theta = \left( {\frac{{10\;{\rm{mm}}}}{{10\;{\rm{mm}} + 7\;{\rm{mm}}}}} \right)\\ \theta = 36.03^\circ \end{array}\]The free body diagram of the picture is shown below:
The force on the member $B$ and $C$ are identical because of the symmetry.
To find the normal force we will equate the sum of forces in $x$ direction by using relation,
\[\begin{array}{c} \Sigma {F_x} = 0\\ {F_A} - {F_B}\cos \theta - {F_C}\cos \theta = 0 \end{array}\]On plugging the values in the above relation, we get,
\[\begin{array}{c} \left( {80\;{\rm{N}}} \right) - {F_B}\cos \left( {36.03^\circ } \right) - {F_B}\cos \left( {36.03} \right)^\circ = 0\\ \left( {80\;{\rm{N}}} \right) - 2{F_B}\cos \left( {36.03} \right)^\circ = 0\\ {F_B} = 49.46\;{\rm{N}} \end{array}\]The normal force on the pipe C is calculated as:
\[{F_B} = {F_C} = 49.46\;{\rm{N}}\]