Question 11.1: Dynamic Force Analysis of a Single Link in Pure Rotation. (S...
Dynamic Force Analysis of a Single Link in Pure Rotation. (See Figure 11‑1)
Given: The 10-in-long link shown weighs 4 lb. Its CG is on the line of centers at the 5-in point. Its mass moment of inertia about its CG is 0.08 lb-in-sec². Its kinematic data are:
\begin{array}{cccc}\theta_{2} deg & \omega_{2} rad / sec & \alpha_{2} rad / sec ^{2} & a_{G_{2}} in / sec ^{2} \\30 & 20 & 15 & 2001 @ 208^{\circ}\end{array}
An external force of 40 lb at 0° is applied at point P.
Find: The force F _{12} \text { at pin joint } O_{2} \text { and the driving torque } T _{12} needed to maintain motion with the given acceleration for this instantaneous position of the link.
1 Convert the given weight to proper mass units, in this case blobs:
\text { mass }=\frac{\text { weight }}{g}=\frac{4 lb }{386 in / sec ^{2}}=0.0104 \text { blob } (a)
2 Set up a local coordinate system at the CG of the link and draw all applicable vectors acting on the system as shown in the figure. Draw a free-body diagram as shown.
3 Calculate the x and y components of the position vectors R _{12} \text { and } R _{P} in this coordinate system:
\begin{array}{lll}R _{12}=5 \text { in @ } \angle 210^{\circ} ; & R_{12_{x}}=-4.33, & R_{12_{y}}=-2.50 \\R _{P}=5 \text { in @ } \angle 30^{\circ} ; & R_{P_{x}}=+4.33, & R_{P_{y}}=+2.50\end{array} (b)
4 Calculate the x and y components of the acceleration of the CG in this coordinate system:
a _{G}=2001 @ \angle 208^{\circ} ; \quad a_{G_{x}}=-1766.78, \quad a_{G_{y}}=-939.41 (c)
Calculate the x and y components of the external force at P in this coordinate system:
F _{P}=40 @ \angle 0^{\circ} ; \quad F_{P_{x}}=40, \quad F_{P_{y}}=0 (d)
Substitute these given and calculated values into the matrix equation 11.4:
\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\-R_{12_{y}} & R_{12_{x}} & 1\end{array}\right] \times\left[\begin{array}{l}F_{12_{x}} \\F_{12_{y}} \\T_{12}\end{array}\right]=\left[\begin{array}{l}m_{2} a_{G_{x}}-F_{P_{x}} \\m_{2} a_{G_{y}}-F_{P_{y}} \\I_{G} \alpha-\left(R_{P_{x}} F_{P_{y}}-R_{P_{y}} F_{P_{x}}\right)\end{array}\right] (11.4)
\begin{gathered}{\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\2.50 & -4.33 & 1\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\T_{12}\end{array}\right]=\left[\begin{array}{c}(0.01)(-1766.78)-40 \\(0.01)(-939.41)-0 \\(0.08)(15)-\{(4.33)(0)-(2.5)(40)\}\end{array}\right]} \\{\left[\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\2.50 & -4.33 & 1\end{array}\right] \times\left[\begin{array}{c}F_{12_{x}} \\F_{12_{y}} \\T_{12}\end{array}\right]=\left[\begin{array}{c}-57.67 \\-9.39 \\101.2\end{array}\right]}\end{gathered} (e)
7 Solve this system either by inverting matrix A and premultiplying that inverse times matrix C using a pocket calculator with matrix capability; using Mathcad or Matlab; or by putting the values for matrices A and C into program Matrix downloadable with this text.
Program Matrix gives the following solution:
F_{12 x}=-57.67 lb , \quad F_{12 y}=-9.39 lb , \quad T_{12}=204.72 lb – in (f)
Converting the force to polar coordinates:
F _{12}=58.43 @ \angle 189.25^{\circ} (g)