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The square frame consists of four identical homogeneous rods of mass $m$. It lies in the vertical plane and can move in it due to the wheels situated in A and B. These wheels can slide frictionlessly along the "white" horizontal and vertical tracks respectively. The frame is released from rest in the position as depicted above. The task is to determine the instantaneous reactions forces in A and B precisely when the frame is released.


Since the tracks are frictionless the (net) reaction force in B is strictly horizontal and in A strictly vertical. Below is my free-body diagram.


(source: draw.to)

Let's begin with $\textbf{F} = m \boldsymbol{a}_{\text{G}}$. We have

$$ m \boldsymbol{a}_{\text{G}} = (\text{N}_{\text{A}} - 4mg) \textbf{e}_y + \text{N}_{\text{B}} \textbf{e}_x \ . $$

Next we will use moment of inertia about an axis through the center of mass $\text{G}$, parallell to the $z$-axis (coming out from the screen). The moment of inertia about $G$ for the entire frame is $I_G = 16ml^2/3$. The torque then is

$$ I_G \dot{\omega} = l ( N_A - N_B) \ , $$

which will give us an expression for the angular acceleration (note that $\omega = 0$ in the instance of release).


This is unfortunately all the progress I've been able to make and I am really clueless as to how I should proceed from here. I think the equation for accelerations in a rigid body make come in handy

$$ \boldsymbol{a}_P = \boldsymbol{a}_Q + \dot{\omega} \textbf{e}_z \times \textbf{r}_{QP} - \underbrace{\omega^2}_0 \textbf{e}_z \ , $$

where $P$ and $Q$ are two points on the rigid body; the question is which points to choose judiciously. $G$ definitely is one of them but I have no information on the acceleration of any other point of the frame.


The answers:

$$ \textbf{N}_A = \frac {14}5 mg \textbf{e}_y , \quad \textbf{N}_B = \frac 65 mg \textbf{e}_x $$

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  • $\begingroup$ This is a near problem. Converting the two sliders (two degrees of freedom) into a single degree of freedom problem is the key to solving it. $\endgroup$ – ja72 Dec 14 '14 at 1:10
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One thing you need to consider is the kinematics. That is what is the position, velocity and acceleration of the center of mass G as the frame rotates. I set the coordinate system origin where the two slots intersect to get

$$\begin{aligned} \vec{r}_A & = (x_A,0,0) & \vec{r}_B & = (0,y_B,0) \end{aligned} $$

The I recognized that the two positions are related by a rigid body transformation

$$ \vec{r}_B = \vec{r}_A + \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} - 2\ell \\ 2 \ell \\ 0 \end{pmatrix} $$

which gives $x_A = 2 \ell (\cos \theta+\sin \theta)$ and $y_B = 2 \ell (\cos \theta-\sin \theta)$. So now we have the positions A, B, and G as a function of the rotation angle $\theta$.

$$\begin{aligned} \vec{r}_A & = (2 \ell (\cos \theta+\sin \theta),0,0) \\ \vec{r}_B & = (0,2 \ell (\cos \theta-\sin \theta),0) \\ \vec{r}_G & = (\ell (\cos \theta+\sin \theta), \ell (\cos \theta-\sin \theta),0) \end{aligned} $$

More importantly we have the influence of the angle, which by using chain rule we get

$$\begin{aligned} \vec{v}_A =(\dot{x}_A,0,0) & = (y_B \dot{\theta},0,0) \\ \vec{v}_B =(0,\dot{y}_B,0) & = (0,-x_A \dot{\theta} ,0) \\ \end{aligned} $$

The velocity kinematics of the center of mass follow the rigid body rule

$$\begin{aligned} \vec{v}_G & = \vec{v}_A + \vec{\omega} \times (\vec{r}_G - \vec{r}_A) \\ \vec{v}_G =(\dot{x}_G,\dot{y}_G,0) & = (\frac{y_B}{2} \dot{\theta},-\frac{x_A}{2} \dot{\theta} ,0) \\ \vec{\omega} & = (0,0,\dot{\theta}) \end{aligned} $$

and

$$\begin{aligned} \vec{a}_A =(\ddot{x}_A,0,0) & = (y_B \ddot{\theta}-x_A \dot{\theta}^2,0,0) \\ \vec{a}_B =(0,\ddot{y}_B,0) & = (0,-x_A \ddot{\theta}-y_A \dot{\theta}^2 ,0) \\ \end{aligned} $$

The acceleration kinematics of the center of mass follow the rigid body rule

$$\begin{aligned} \vec{a}_G & = \vec{a}_A +\dot{\vec{\omega}}\times (\vec{r}_G - \vec{r}_A)+ \vec{\omega} \times (\vec{v}_G - \vec{v}_A) \\ \vec{a}_G =(\ddot{x}_G,\ddot{y}_G,0) & = (\frac{y_B}{2} \ddot{\theta}-\frac{x_A}{2} \dot{\theta}^2,-\frac{x_A}{2} \ddot{\theta}+\frac{y_B}{2} \dot{\theta}^2 ,0) \end{aligned} $$

Now we can look at sum of forces and moments about G

$$\begin{aligned} \vec{N}_A + \vec{N}_B & = 4 m \vec{a}_G \\ (\vec{r}_A-\vec{r}_G) \times \vec{N}_A +(\vec{r}_B-\vec{r}_G) \times \vec{N}_B & = I \dot{\vec{\omega}} \end{aligned} $$

which expanded out by component yields the following system of 3 equations.

$$\begin{aligned} N_B & = 4 m \ddot{x}_G =4 m \left(\frac{y_B}{2} \ddot{\theta}-\frac{x_A}{2} \dot{\theta}^2\right) \\ N_A-4 m g & =4 m \ddot{y}_G =4 m \left(-\frac{x_A}{2} \ddot{\theta}+\frac{y_B}{2} \dot{\theta}^2 \right)\\ N_A \frac{x_A}{2} - N_B \frac{y_A}{2} & = I_z \ddot{\theta} \end{aligned} $$

where $I_z = \frac{16}{3} m \ell^2$

The answer is found when $\theta=0$, $\dot\theta=0$, $x_A=2 \ell$ and $y_B=2 \ell$ since the above is 3 equations with 3 unknowns ($N_A$, $N_B$ and $\ddot{\theta}$). The solution I get is

$$ \begin{aligned} N_A & = \frac{g 4 m (I_z+4 m \ell^2)}{I_z+8 m \ell^2} = \frac{14 m g}{5}\\ N_B & = \frac{g 16 m^2 \ell^2}{I_z + 8 m \ell^2} = \frac{6 m g}{5} \\ \ddot{\theta} &= \frac{g 4 m \ell}{I_z + 8 m \ell^2} = \frac{3 g}{10 \ell} \end{aligned} $$

The answer here what you expect.

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  • $\begingroup$ Sorry - no. The answer key is correct. You made a mistake. $\endgroup$ – Floris Dec 13 '14 at 23:53
  • $\begingroup$ What I did wrong is $m$ which for me is the total mass of the 4 rods. Hence the $I_z$ and $m$ is not correct above. $\endgroup$ – ja72 Dec 14 '14 at 0:45
  • $\begingroup$ That doesn't explain the discrepancy. $F_B=\frac37 F_A$ regardless of the mass. You get $F_B=\frac{3}{19}F_A$... there's something else going on. $\endgroup$ – Floris Dec 14 '14 at 0:51
  • $\begingroup$ I have updated the answer. The method was correct, I just assumed a different mass variable. $\endgroup$ – ja72 Dec 14 '14 at 0:52
  • $\begingroup$ It is sufficient to just change $m \rightarrow 4m$ you also need to have consistent $I_z$. You need to make the substitution before the value of $I_z = \frac{16}{3} m \ell^2$ is used since it contains the rod mass, and not the net mass. $\endgroup$ – ja72 Dec 14 '14 at 0:52
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Draw the position of the frame a very small time later. From this you can determine the relative rotation (due to torque - the difference between the forces multiplied by their perpendicular distance to the center of gravity) and linear acceleration (due to net horizontal and vertical forces). That should give you all the equations you need and you can then solve for $F_A$ and $F_B$.

A few hours later...

Here is the diagram I had in mind - drawing a "large" $\theta$ because it's easier to see what is going on:

enter image description here

The distance $D=L\sqrt{2}$ is a convenient quantity to help compute the new position (x,y) at time t.

Next we need to calculate the moment of inertia of the square: from the parallel axis theorem, we find it is

$$I = 4(\frac{1}{12}m(2L)^2 + mL^2)=\frac{16}{3} mL^2$$

Now we have the following equations of motion:

$$4m\ddot{x}=F_B\\ 4m\ddot{y}=4mg-F_A\\ I\ddot{\theta}=(F_B-F_A)L$$

And finally, when $\theta=\frac{\pi}{4}$ (start of motion) we can write

$$\frac{dy}{d\theta}=D\cos(\frac{\pi}{4}) = L\\ \frac{dx}{d\theta}=-D\sin(\frac{\pi}{4}) = -L$$

This means

$$\ddot{x}=-L\ddot{\theta}\\ \ddot{y}=L\ddot{\theta}$$

We can now eliminate $\ddot{x}$ and $\ddot{y}$ and are left with three equations in three unknowns: $$4mL\ddot{\theta}=-F_B \tag1$$ $$4mL\ddot{\theta}=4mg-F_A \tag2$$ $$\frac{16}{3} m L \ddot{\theta} = F_A-F_B \tag3$$

Subtracting $(1)$ and $(2)$ gives

$$4mg=F_A+F_B\tag{4}$$

and $(1) - \frac34 (3)$ gives

$$F_B=\frac34(F_A-F_B)$$ $$F_B=\frac37 F_A\tag{5}$$

$(5)$ into $(4)$ gives

$$4mg = \frac{10}{7} F_A\\ F_A = \frac{14}{5} mg$$

and then it follows that

$$F_B = \frac65 mg$$

Just as your answer key showed.

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  • $\begingroup$ Hi. "Draw the position of the frame a very small time later", I can't say I'm sure how to interpret. Infinitesimally small time later or enough time so it looks slanted? Also I can't say I'm fully aware what further information/equations this provides. Could you give some further hints? Thanks for the help! $\endgroup$ – larrydavid Dec 13 '14 at 21:26
  • $\begingroup$ The only thing confusing for me is that a positive $\theta$ is a clockwise rotation which means that a negative torque is needed to increase $\theta$. This makes it difficult to ensure consistency. $\endgroup$ – ja72 Dec 14 '14 at 1:09
  • $\begingroup$ It was convenient to draw that particular angle. Sorry it was confusing! $\endgroup$ – Floris Dec 14 '14 at 3:48
  • $\begingroup$ Why is not the center of mass force 4mg considered in the torque equation $I \ddot{\theta}$ ? $\endgroup$ – larrydavid Dec 16 '14 at 18:09
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    $\begingroup$ Ah ok! How subtle indeed that $\theta$ was unconventionally oriented. Quite complex and unintuitive stuff to digest here, I'll have to give it some time and thought. Thanks for all help $\endgroup$ – larrydavid Dec 16 '14 at 18:49

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