Determining Friction and Contact forces

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.

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

• Are the parameters for the problem like $s, b, X_A$ the same like your previous question?. Because as you are stating the problem it is not possible to calculate numerically the tangent at B. Commented Nov 23, 2020 at 10:23
• Its the same equations, but in the previous question that wasn't anything that was ever covered. The s,b, and Xa weren't even needed to solve the previous question. Commented Nov 23, 2020 at 10:45
• you'd s,b, Xa and c need them to calculate the angle at B. Commented Nov 23, 2020 at 13:37
• At that point I wouldn't know how to do that. Last time I even saw that was in trig which was 10 years ago over 6 weeks during the summer, while working a graveyard shift and another fulltime job while battling a severe sinus infection for the final 3 weeks. The example I have doesn't follow that, however it's for a 2 bar element instead of 1 bar and doesn't include having to calculate a seperate angle. Commented Nov 24, 2020 at 2:40
• @BarrettCloud $s, \, b$ and $X_a$ are very much needed for the answer of your previous question, as well as for this one. You should look at what I wrote as an answer. The solution I proposed is based on a method from calculus called related rates. Commented Nov 24, 2020 at 21:16

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.