# 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. – NMech Nov 23 '20 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. – Barrett Cloud Nov 23 '20 at 10:45
• you'd s,b, Xa and c need them to calculate the angle at B. – NMech Nov 23 '20 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. – Barrett Cloud Nov 24 '20 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. – Futurologist Nov 24 '20 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.