Acceleration Analysis

Significance

This is important because of $\Sigma F=ma$ where we want to find points of large force

For solving questions, you need to be given ${\theta }_2$, ${\omega }_2$, and ${\alpha }_2$

Acceleration of a point P Polar Form $A_p=p\alpha j{e}^{j\theta }-p{\omega }^2{e}^{j\theta }$
Rectangular Form
$A_p=p\alpha \left(-\sin \theta +j\cos \theta \right)-p{\omega }^2\left(\cos \theta +j\sin \theta \right)$
Where the first term is the tangential component and the second term is the normal component

Types of acceleration

Absolute Acceleration: Acceleration w.r.t a point on the same body fixed at the origin of a global coordinate system
Acceleration Difference: Acceleration with respect to a point on the same body that is not fixed
Relative Acceleration: Acceleration with respect to a point on a different body that is not fixed

Vector Loop Analysis For a Four Bar Linkage

1. Check the reference frame / coordinate system to make sure that $R_1$ is fixed with an angle of zero w.r.t the x-axis
2. If you don’t have ${\theta }_3$ and ${\theta }_4$ you need to do a position analysis first, if you don’t have ${\omega }_3$ and ${\omega }_4$ you need to do a velocity analysis first
3. Solve for the ABC’s

$$A=c\sin {\theta }_4$$ $$B=b\sin {\theta }_3$$ $$C=a{\alpha }_2\sin {\theta }_2+a{\omega }_2^2\cos {\theta }_2+b{\omega }_3^2\cos {\theta }_3-c{\omega }_4^2\cos {\theta }_4$$ $$D=c\cos {\theta }_4$$ $$E=b\cos {\theta }_3$$ $$F=a{\alpha }_2\cos {\theta }_2-a{\omega }_2^2\sin {\theta }_2-b{\omega }_3^2\sin {\theta }_3+c{\omega }_4^2\sin {\theta }_4$$

4. Calculate angular accelerations

$${\alpha }_3=\frac{CD-AF}{AE-BD}$$ $${\alpha }_4=\frac{CE-BF}{AE-BD}$$

5. Solve for linear accelerations in potentially both configurations (open and crossed)

$${\mathbf{A}}_A=a{\alpha }_2\left(-\sin {\theta }_2+j\cos {\theta }_2\right)-a{\omega }_2^2\left(\cos {\theta }_2+j\sin {\theta }_2\right)$$ $$=-a{\alpha }_2\sin {\theta }_2-a{\omega }_2^2\cos {\theta }_2+j\left(a{\alpha }_2\cos {\theta }_2-a{\omega }_2^2\sin {\theta }_2\right)$$ $${\mathbf{A}}_{BA}=b{\alpha }_3\left(-\sin {\theta }_3+j\cos {\theta }_3\right)-b{\omega }_3^2\left(\cos {\theta }_3+j\sin {\theta }_3\right)$$ $$=-b{\alpha }_3\sin {\theta }_3-b{\omega }_3^2\cos {\theta }_3+j\left(b{\alpha }_3\cos {\theta }_3-b{\omega }_3^2\sin {\theta }_3\right)$$ $${\mathbf{A}}_B=c{\alpha }_4\left(-\sin {\theta }_4+j\cos {\theta }_4\right)-c{\omega }_4^2\left(\cos {\theta }_4+j\sin {\theta }_4\right)$$ $$=-c{\alpha }_4\sin {\theta }_4-c{\omega }_4^2\cos {\theta }_4+j\left(c{\alpha }_4\cos {\theta }_4-c{\omega }_4^2\sin {\theta }_4\right)$$

Vector Loop Analysis For a Cranker-Slider

Once again, make sure $R_1$ is coincident with the X-axis

Parameter check?

If you don’t have ${\theta }_3$ you need to do a position analysis first

If you don’t have ${\omega }_3$ and $\dot{d}$ you need to do a velocity analysis first

Angular Acceleration and linear acceleration of the slider block
${\alpha }_3=\frac{a{\alpha }_2\cos {\theta }_2-a{\omega }_2^2\sin {\theta }_2+b{\omega }_3^2\sin {\theta }_3}{b\cos {\theta }_3}$
$d^{{}^{\prime \prime }}=-a{\alpha }_2\sin {\theta }_2-a{\omega }_2^2\cos {\theta }_2+b{\alpha }_3\sin {\theta }_3+b{\omega }_3^2\cos {\theta }_3$
Linear Accelerations

$${\mathbf{A}}_A=a{\alpha }_2\left(-\sin {\theta }_2+j\cos {\theta }_2\right)-a{\omega }_2^2\left(\cos {\theta }_2+j\sin {\theta }_2\right)$$ $$=-a{\alpha }_2\sin {\theta }_2-a{\omega }_2^2\cos {\theta }_2+j\left(a{\alpha }_2\cos {\theta }_2-a{\omega }_2^2\sin {\theta }_2\right)$$ $${\mathbf{A}}_{BA}=b{\alpha }_3\left(-\sin {\theta }_3+j\cos {\theta }_3\right)-b{\omega }_3^2\left(\cos {\theta }_3+j\sin {\theta }_3\right)$$ $$=-b{\alpha }_3\sin {\theta }_3-b{\omega }_3^2\cos {\theta }_3+j\left(b{\alpha }_3\cos {\theta }_3-b{\omega }_3^2\sin {\theta }_3\right)$$ $${\mathbf{A}}_B=-a{\alpha }_2\sin ({\theta }_2)-a{\omega }_2^2\cos ({\theta }_2)+b{\alpha }_3\sin ({\theta }_3)+b{\omega }_3^2\cos ({\theta }_3)$$

Note: You may need to solve twice for the open and crossed configurations

Sign notation for the slider

Note that a + $A_B$ for the slider means downwards.