Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 99P from Chapter 17 from Hibbeler's Engineering Mechanics.
Problem 99P
Step-by-Step Solution
We are given the mass, horizontal force and length of bar as $m = 12\;{\rm{kg}}$, $F = 80\;{\rm{N}}$ and $L = 2\;{\rm{m}}$ respectively.
We are asked to determine the acceleration of the center of the roller at the instant the force is applied.
Step 2
The following is the free body diagram.
To find the forces in x-direction, we will use the relation,
\[\begin{array}{c} \Sigma {F_x} = ma\\ \left( {80\;{\rm{N}}} \right) = \left( {12\;{\rm{kg}}} \right)a\\ a = 6.66\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]Step 3
To find the angular acceleration, we will use the relation,
\[\begin{array}{c} \Sigma {M_A} = {M_0}\\ \left( {ma \times h} \right) = \left( {I\alpha } \right)\\ \left( {ma \times h} \right) = \left( {\frac{1}{{12}}m{L^2}} \right)\alpha \end{array}\]Here, h is the height of the bar from point A to G.
Step 4
On plugging the values in the above relation, we get,
\[\begin{array}{c} \left( {12\;{\rm{kg}} \times 6.66\;{\rm{m/}}{{\rm{s}}^2} \times 1\,{\rm{m}}} \right) = \left( {\frac{1}{{12}}\left( {12\;{\rm{kg}}} \right){{\left( {2\;{\rm{m}}} \right)}^2}} \right)\alpha \\ \alpha = 19.98\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]To find the acceleration of the center of the roller, we will use the relation,
\[\begin{array}{c} \vec a = {{\vec a}_A} + \left( {\alpha \times h} \right) - {\omega ^2}r\\ ai + a'j = {a_A}i + \left( { - 19.98k \times 1j} \right) - 0\\ 6.66i + a'j = \left( {{a_A} - \left( {19.98} \right)} \right)i \end{array}\]Step 5
On equating the components of $i$, we get,
\[\begin{array}{c} 6.66 = {a_A} - 19.98\\ {a_A} = 26.64\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]On equating the components of $j$, we get,
\[a' = 0\]