Question 15.8: The linkage ABDE moves in the vertical plane. Knowing that i......
The linkage ABDE moves in the vertical plane. Knowing that in the position shown crank AB has a constant angular velocity \text V_1 of 20 rad/s counterclockwise, determine the angular velocities and angular accelerations of the connecting rod BD and of the crank DE.
This problem could be solved by the method used in Sample Prob. 15.7. In this case, however, the vector approach will be used. The position vectors {r}_{B}, {r}_{D},~ \text{and}~ {r}_{D / B} are chosen as shown in the sketch.
Velocities. Since the motion of each element of the linkage is contained in the plane of the figure, we have
\text{V}_{A B}=\text{v}_{A B} {k}=(20 ~\mathrm{rad} / \mathrm{s}) {k} \quad\quad \text{V}_{B D}=\text{v}_{B D} {k} \quad\quad \text{V}_{D E}=\text{v}_{D E} \mathbf{k}where k is a unit vector pointing out of the paper. We now write
Equating the coefficients of the unit vectors i and j, we obtain the following two scalar equations:
\begin{aligned}& -17 \text{v}_{D E}=-280-3 \text{v}_{B D} \\& -17 \text{v}_{D E}=+160+12 \text{v}_{B D} \\& \mathrm{~V}_{B D}=-(29.33 ~\mathrm{rad} / \mathrm{s}) {k} \quad\quad \mathrm{V}_{D E}=(11.29~\mathrm{rad} / \mathrm{s}){k}\end{aligned}Accelerations. Noting that at the instant considered crank AB has a constant angular velocity, we write
\mathrm{A}_{A B}=0 \quad\quad \mathrm{~A}_{B D}=\mathrm{a}_{B D} {k} \quad\quad \mathrm{A}_{D E}=\mathrm{a}_{D E} {k} \\ {a}_{D}={a}_{B}+{a}_{D / B} (1)
Each term of Eq. (1) is evaluated separately:
a _D= a _{D E} k \times r _D-\text v _{D E}^2 r _D \\ \\= a _{D E} k \times(-17 i +17 j )-(11.29)^2(-17 i +17 j ) \\ \\ =-17 a _{D E} j -17 a _{D E} i +2170 i -2170 ja _B= a _{A B} k \times r _B- \text v _{A B}^2 r _B=0-(20)^2(8 i +14 j )
= -3200i – 5600j
a _{D / B}= a _{B D} k \times r _{D / B}-\text v _{B D}^2 r _{D / B} \\ \\ = a _{B D} k \times(12 i +3 j )-(29.33)^2(12 i +3 j ) \\ \\ =12 a _{B D} j -3 a _{B D} i -10,320 i -2580 jSubstituting into Eq. (1) and equating the coefficients of i and j, we obtain
\begin{aligned}& -17 \mathrm{a}_{D E}+3 \mathrm{a}_{B D}=-15,690 \\& -17 \mathrm{a}_{D E}-12 \mathrm{a}_{B D}=-6010 \\& \mathrm{~A}_{B D}=-\left(645~ \mathrm{rad} / \mathrm{s}^{2}\right) {k} \quad\quad \mathrm{A}_{D E}=\left(809~ \mathrm{rad} / \mathrm{s}^{2}\right) {k}\end{aligned}