Determining Friction and Contact forces

Question

I am looking at a rod that has an acting force along one end that's pinned into a frictionless track. The other end is pinned into a track that has friction. I am trying to determine the contact forces at each end of the rod, as well as the friction force on the end that's in the frictioned track.

$$ \vec{\dot{\omega}} = 0.1 $$

I am looking at using the equations $ M_{cm} = I_{cm} $ and $ F = mA_{cm} $

I've worked the force equations to equal $$ F_x = mA_{cm} $$ $$ 5-cos(\phi)F_b = sin(27.65)98.1N $$

and

$$ F_y = mA_{cm} $$ $$ N_A + sin(\gamma)N_B+cos(\phi)=cos(27.65)*98.1 $$

and the momentum equation

$$ M_{cm} = I_cm * \vec{\dot{\omega}} $$ $$ M_{cm} = 10 * 0.1 $$

I'm having difficulties figuring out the angle for $ F_B$ as well as determining how $M_{cm}$ is playing a role in here at all.

Any suggestions would be appreciated.

$$ I = 1.41 $$ $$ s = 0.5 $$ $$ b = -0.5 $$ $$r = 1$$ $$c = 2$$ $$X_A = 1.25$$ $$V_A = .5$$ $$A_A= 0.25 $$

Answer 1

Maybe something like this:

Step 1: By the previous answer, the coordinate $X_A = X_A(t)$ and the angle $\theta = \theta(t)$ are connected by the equation $$\big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)^2 \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)^2 \, = \, r^2$$ Knowing $X_A$, plug it in the equation and solve for $\theta$.

Step 2: Differentiate the equation from step 1 with respect to $t$ and obtain the equation $$\big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)\left(\frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)\left(s\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt}\right) = 0$$ Knowing $X_A, \, \theta, \, \frac{dX_A}{dt} = V_A$, plug them in the equation and solve for $\frac{d\theta}{dt}$.

Step 3: Differentiate the equation from step 2 with respect to $t$ and obtain the equation (which you should calcualte): $$\frac{d}{dt} \left( \, \big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)\left(\frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)\left(s\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, \right) = 0$$ for $\frac{d\theta}{dt}$. This will be an equation for $X_A, \theta, V_A = \frac{dX_A}{dt}, \, \frac{d\theta}{dt}, \, A_A = \frac{d^2X_A}{dt^2}$ and $\frac{d^2\theta}{dt^2}$. Given $X_A, \theta, V_A = \frac{dX_A}{dt}, \, \frac{d\theta}{dt}, \, A_A = \frac{d^2X_A}{dt^2}$, plug them in the equation and solve for $\frac{d^2\theta}{dt^2}$.

Step 4: Calculate the coordinates of the point $B$ on the circle: \begin{align} &X_B = X_A + l\cos(\theta + \varphi_0) \\ &Y_B = sX_A + l\sin(\theta + \varphi_0) + b \end{align}

Step 5: Calculate the coordinates of the unit vectors $\vec{n}_A$ and $\vec{n}_B$ at points $A$ and $B$, perpendicular to the line and the circle respectively. For $\vec{n}_A$: \begin{align} &n_{x,A} = \frac{-s}{\sqrt{1+s^2}} \\ &n_{y,A} = \frac{1}{\sqrt{1+s^2}} \end{align} and for $\vec{n}_B$: \begin{align} &n_{x,B} = \frac{X_B - c}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{X_B - c}{r}\\ &n_{y,B} = \frac{Y_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{Y_B}{r} \end{align}

Step 6: Calculate the coordinates of the unit vectors $\vec{t}_A$ and $\vec{t}_B$ at points $A$ and $B$, tangent to the line and the circle respectively. For $\vec{t}_A$: \begin{align} &t_{x,A} = \frac{1}{\sqrt{1+s^2}} \\ &t_{y,A} = \frac{s}{\sqrt{1+s^2}} \end{align} and for $\vec{t}_B$: \begin{align} &t_{x,B} = \frac{Y_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{Y_B}{r}\\ &t_{y,B} = \frac{c - X_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{c - X_B}{r} \end{align}

Step 7: Calculate the acceleration of point $B$, by differentiating twice with respect to time $t$ the equation for the coordinates $X_B, \, Y_B$: \begin{align} &A_{x,B} = \frac{d^2X_A}{dt^2} = A_A - l\sin(\theta + \varphi_0)\frac{d^2\theta}{dt^2} - l\cos(\theta + \varphi_0)\left(\frac{d\theta}{dt}\right)^2\\ &A_{y,B} = \frac{d^2Y_A}{dt^2} =s\,A_A + l\cos(\theta + \varphi_0)\frac{d^2\theta}{dt^2} - l\sin(\theta + \varphi_0)\left(\frac{d\theta}{dt}\right)^2 \end{align} Since, up to know, you are either given or have calculated the variables $A_A, \, \theta, \, \frac{d\theta}{dt}, \, \frac{d^2\theta}{dt^2}$, you can plug them in the equations and calculate the coordinates $A_{x,B}, \, A_{y,B}$ of the acceleration vector $\vec{A}_B$ of point $B$. The coordinates of the acceleration vector $\vec{A}_A$ of point $A$ are $A_{A}, \, sA_{A}$.

Step 8: Finally put together the linear system of three scalar equations and three unknown force magnitudes $N_A, \, N_B, \, F_B$: \begin{align} & \left(\frac{ml^2}{12}\right) \, \frac{d^2\theta }{dt^2} \, \hat{k} \, = \, \frac{1}{2} \, \big(l \, \cos(\theta + \varphi_0)\hat{i} + l \, \sin(\theta + \varphi_0)\hat{j}\big) \times \left(\,F_B \vec{t}_B + N_B\vec{n}_B - P_A \vec{t}_A - N_A \vec{n}_A\,\right)\\ &\frac{m}{2} \big(\vec{A}_A + \vec{A}_B\big) = F_B \vec{t}_B + N_B\vec{n}_B + P_A \vec{t}_A + N_A \vec{n}_A \end{align} and since you already know all the other parameters in this system, including $P_A$ which is given, solve for $N_A, \, N_B, \, F_B$. Here $\hat{i}, \, \hat{j}, \, \hat{k}$ are the pairwise orthogonal unit vectors of the inertial coordinate system of the system. The vector $\hat{k}$ is perpendicular to the 2D picture.