Find frictional losses as box exits ramp

I have a box traveling on a ramp just before the ramp levels out. At this point all values are known: I want to calculate the frictional losses from this point to when the box is completely off the ramp and level. The picture below shows the box in the middle of this process and the forces I think that apply: The problem I am facing is that the normal forces and frictional reaction forces change depending on where the box is during this process. I don't know how to write an equation that accounts for this. I am assuming this is a completely inelastic collision and that all energy is transferred horizontally.

I feel that some calculus is needed, and that an equation for both contact points would be in respect to the angle AND either the displacement along the x axis or ramp surface. I need help in creating these equations.

• one question for you. Are either the ramp or the horizontal surface moving ( i.e. are they some kind of conveyor belt)? Nov 30 '20 at 9:43
• The box on top is moving (the ramp has roller bearings). But the ramp and round are stationary. Nov 30 '20 at 20:09

Warning: what you are setting out to do is achievable, but the calculations are quite bothersome. I will offer a static approximation for the reaction forces, and then the outline of the dynamic "proper" solution. Once you have the reaction forces with respect to a given angle(and some other parameters if you choose the hard way), you can iterate for small timesteps to obtain the solution (like you suggested).

General intro

Let's use the points O, A, C in the image below. Also please notice that I corrected the the normal reaction to point A and the direction of friction. The friction is opposing motion, and the motion is along the ramp. Finally, if $$\phi$$ is the angle that the bottom side of the box forms with the horizontal plane, then you can find the coordinates of the point A, C with respect to angle $$\phi$$, by using the rotation matrix

$$R(\phi) =\begin{bmatrix}\cos(\phi)& -\sin(\phi) & 0 \\ \sin(\phi)&\cos(\phi)& 0 \\0 & 0 &1\end{bmatrix}$$

For example for point A, the coordinate when the box is completely flat with respect to O would be $$A_0=[l_p, 0,0]^T$$ (from now on, for simplicity, i'll be lazy and skip the formal notation $$[]^T$$ for simplicity and just use $$A_0=[l_p, 0,0]$$ for a vector).

This will yield that the coordinates of A wrt to $$\phi$$ are:

$$A(a) = R(a)A_0 = [l_p\cos(\phi), l_p\sin(\phi),0]$$

Similarly you can calculate the coordinates for C with respect to point O. $$C(a) = R(\phi)C_0 = \left[\frac{l_p}{2}\cos(\phi)- \frac{h_p}{2}\sin(\phi), \frac{l_p}{2}\sin(\phi) + \frac{h_p}{2}\cos(\phi)\right]$$

Now you can easily write the equations of dynamic equilibrium to calculate reaction forces and accelerations.

$$\sum F_x = m a_{C,x} \rightarrow \mu_c F_{n,c} + \mu_r F_{n,r}\cos(\theta) - F_{n,r}\sin(\theta) =m a_{C,x}$$ $$\sum F_y = m a_{C,y} \rightarrow F_{n,c} + \mu_r F_{n,r}\sin(\theta) + F_{n,r}\cos(\theta) - F_g =m a_{C,y}$$ $$\sum M_O = I \ddot{a} \rightarrow \vec{r}_{c} \times \vec{F}_{g} + \vec{r}_{OA} \times\left(\overrightarrow{\mu_r F_{n,r}} + \vec{F}_{n,r}\right) =I \ddot{\phi} + r_{OC}\times (m \vec{a}_C)$$

Again, I am being very lazy and I:

• denote $$\overrightarrow{\mu_r F_{n,r}}$$ for the vector of the friction force on the ramp
• use the vector form for a force moment $$\vec{r}\times\vec{F}$$, which is a shorthand for: $$\vec{r}\times\vec{F}= \left|\begin{matrix}\vec{i} &\vec{j}&\vec{k} \\ r_x & r_y & r_z\\ F_x & F_y & F_z \end{matrix}\right|$$

STATIC approximation

At this point, there is no appreciable acceleration, you can get away with setting $$a_{C,x},a_{C,y}, \ddot{\phi}$$ equal to zero and you just solve the system and estimate the reaction forces.

That will be a lot easier, and you'd be within 10% error (if the assumptions are valid).

If that's not good enough, then proceed to the next section( at your own peril :-) ).

Truly Dynamic

AS you might have noticed you now have 3 equations with (Seemingly) 5 unknowns ($$F_{n,c}, F_{n,r}, a_{C,x},a_{C,y}, \ddot{\phi}$$). The word Seemingly is stressed because you also need to know $$\dot{\phi}$$ which doesn't come into play yet but it will in a moment.

So in order to solve the system you'd need at least another 3 equations. What you can use is the relative velocity and acceleration between points O,and A. Basically, the velocity and acceleration at O should only have a horizontal component, while similarly the components velocity and acceleration of point A should comply with the following constraint $$\tan\theta=\frac{a_{A,y}}{a_{A,x}}$$.

So for example, the relative velocity of A with respect to O is:

$$\begin{cases} \vec{v}_A = \vec{v}_O + \dot{\phi}\times \vec{r}_{AO} \end{cases}$$ which - considering $$\tan\theta=\frac{v_{A,y}}{v_{A,x}}$$ - can be expanded to $$\begin{cases} v_{A,x} = v_{O,x} + \left(\dot{\phi}\times \vec{r}_{OA}\right)_x \\ v_{A,y} = v_{O,y} + \left(\dot{\phi}\times \vec{r}_{OA}\right)_y \end{cases} \rightarrow \begin{cases} v_{A,x} = v_{O,x} + \left(\dot{\phi}\times \vec{r}_{OA}\right)_x \\ v_{A,x}\tan(\theta) = \left(\dot{\phi}\times \vec{r}_{OA}\right)_y \end{cases}$$

So in the above equation you can estimate the $$\dot{\phi}$$ if you know the velocity of $$v_{O,x}$$. However $$v_{O,x}$$ should be known or can be obtained if you know the kinetic energy of the system at any point (you'd need to calculate first the magnitude of $$v_C$$, and then use a similar equation to relate $$v_A$$ to $$v_C$$. In the simplest case you can start from velocity equal to 0.

The same procedure can be done for the acceleration:

$$\vec{a}_A = \vec{A}_O + \vec{a}_{A|O}= \vec{A}_O + \vec{\dot{\phi}}\times(\vec{\dot{\phi}}\times \vec{r}_{OA}) + \vec{\ddot{\phi}}\times r_{OA}$$

The point here is to calculate $$\ddot{\phi}$$, which can be done in a similar way to the $$\dot{\phi}$$.

If you got $$\dot{\phi}$$ and $$\ddot{\phi}$$ then you can relate the acceleration of the center of mass to the acceleration at either points A, O.

Now you should have all available equations to solve the problem. The problem is writing them down in a clear and concise manner and then solve the system. .

This is not an answer but to express how would I attack this problem. At any time $$t$$, there is a resistant force at the toe of the moving body, and the motion will continue if it is surpassed. Assume the sketch below is at $$t = 0$$, and $$d = 0$$, your job is then to find the energy accumulated between $$t - t_f$$ and $$d - L$$, in which $$L$$ is the length of the moving body. Hope this helps. 