# Physics Challenge April 2020

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Chestermiller
Mentor
You're right, I concluded that the torques of the ##y## components cancelled but of course they are both in the same sense.
I've decided that I'm done screwing around with this problem. I'm going to solve it only using the laboratory (fixed) frame of reference, and without using the resolution of the problem into two separate problems that I've done so far. To do this, I'm going to focus on the top half of the truss, and do force and moment balances of struts AB and CD.

First the kinematics. There are only two parameters that characterize the kinematics of this problem. This is the angle ##\theta## and the location of the center of mass of the overall rhombus. Using the analysis and notation in post #29, the acceleration components of the center of mass AB are:

##a_C+a_{Ax}/2## in the x direction
and
##-a_{Ax}/2## in the y direction

And for strut BC, they are
##a_C-a_{Ax}/2## in the x-direction
and
##-a_{Ax}/2## in the y direction##
where ##a_C## is the x-direction acceleration of the center of mass of the overall truss.

The figure below shows the free body diagram for the two struts comprising the upper half of the truss:

Based on this, the force- and moment balances on strut AB are:
$$\frac{F}{2}-T=m\left(a_C+\frac{a_{Ax}}{2}\right)$$
$$R_B-R_A=-m\frac{a_{Ax}}{2}$$
$$(R_B+R_A)\frac{b}{2\sqrt{2}}-\left(\frac{F}{2}+T\right)\frac{b}{2\sqrt{2}}=m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$
And for strut BC, they are:
$$T=m\left(a_C-\frac{a_{Ax}}{2}\right)$$
$$R_B+R_C=m\frac{a_{Ax}}{2}$$
$$-(R_C-R_B-T)\frac{b}{2\sqrt{2}}=-m\frac{b^2}{12}\frac{d^2\theta}{dt^2}$$

OK so far?

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JD_PM, TSny and etotheipi
etotheipi
2019 Award
##a_C+a_{Ax}/2## in the x direction
and
##-a_{Ax}/2## in the y direction
And this is because in the CM frame where ##(x,y) = (\frac{b}{2}\cos{\theta}, \frac{b}{2}\sin{\theta})##, we also have ##(\ddot{x}, \ddot{y}) = (-\frac{b}{2}\cos{\theta} (\dot{\theta})^2 - \frac{b}{2}\sin{\theta}\ddot{\theta}, \frac{b}{2}\cos{\theta}\ddot{\theta} - \frac{b}{2}(\dot{\theta})^2 \sin{\theta})##. And initially with ##\theta = \frac{\pi}{4}## that does mean that the initial ##y## acceleration of the CM of AB is ##-\frac{a_{Ax}}{2}##. The torque and force analyses on the struts look good.

I spent a lot of time thinking about your first method yesterday and it's really clever. I came to the conclusion that so long as you can assume that ##\vec{a}_C## is a linear function of all of the applied forces, then you can apply the superposition principle.

Chestermiller
Mentor
And this is because in the CM frame where ##(x,y) = (\frac{b}{2}\cos{\theta}, \frac{b}{2}\sin{\theta})##, we also have ##(\ddot{x}, \ddot{y}) = (-\frac{b}{2}\cos{\theta} (\dot{\theta})^2 - \frac{b}{2}\sin{\theta}\ddot{\theta}, \frac{b}{2}\cos{\theta}\ddot{\theta} - \frac{b}{2}(\dot{\theta})^2 \sin{\theta})##. And initially with ##\theta = \frac{\pi}{4}## that does mean that the initial ##y## acceleration of the CM of AB is ##-\frac{a_{Ax}}{2}##. The torque and force analyses on the struts look good.
OK. Continuing with the analysis, if we add the force- and moment balances for the two struts together, we obtain the following:
$$\frac{F}{2}=2ma_C\tag{1}$$
$$2R_B+R_C-R_A=0\tag{2}$$
$$2R_B+R_A-R_C-\frac{F}{2}=0\tag{3}$$
From Eqn. 1, if follows that $$a_C=\frac{F}{4m}\tag{4a}$$From Eqns. 2 and 3, it follows that $$R_B=\frac{F}{8}\tag{4b}$$
Eqns. 4 are identical to the relationships obtained previously for Part 1 loading. So adding the force and moment balance equations for the two struts is found to deliver the exact same equations as for Part 1 loading.

From Eqn. 2, it follows that $$\frac{(R_A-R_C)}{2}=R_B=\frac{F}{8}\tag{5}$$
Now, with no loss of generality, we can then also write:
$$R_A=R+R_B\tag{6a}$$
$$R_C=R-R_B\tag{6b}$$ where $$R=\frac{(R_A+R_C)}{2}$$
Next, subtracting the force and moment balances of the two struts and combining these with Eqns. 6 yields:
$$\frac{F}{2}-2T=ma_{Ax}\tag{7a}$$
$$2R=ma_{Ax}\tag{7b}$$
$$\left(2R-2T-\frac{F}{2}\right)=2m\frac{b^2}{12}\frac{d^2\theta}{dt^2}\tag{7c}$$

Eqns. 7 are identical to the relationships obtained previously for Part 2 loading. So subtracting the force and moment balance equations for the two trusses is found to deliver the exact same equations as for Part 2 loading.

JD_PM, TSny and etotheipi
etotheipi
2019 Award
Eqns. 7 are identical to the relationships obtained previously for Part 2 loading. So subtracting the force and moment balance equations for the two trusses is found to deliver the exact same equations as for Part 2 loading.
Amazing. You make it look easy! Do you get double points for solving it twice?

It'll take me a little while to put all of the pieces together but I could follow through your reasoning. Thanks!

Chestermiller
My 2 cents worth contribution to this beautiful thread.

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

Where ##O## is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where ##\vec F_i^S## is the specified force at a given point and ##\vec r_i## is the distance from ##O## to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\theta$$

2) Determine the kinetic energy ##(T)## of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame ##XYS##. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it ##T_1##) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it ##T_2##).

- Let's determine ##T_1##: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine ##T_2##: Let's first get the coordinates of the center of mass of the ##AB## rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: ##\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y##.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the ##AB## rod, the squared modulus of the velocity (##|\vec v_{T_2}|^2##) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possess moment of inertia with respect to their center of mass (##\frac{mb^2}{12}## each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus ##T_2## is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy ##(T)## of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For ##x## we get

$$4m \ddot x = F$$

- For ##\theta## we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get ##a_c##

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$

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Spinnor and etotheipi
My 2 cents worth contribution to this beautiful thread.

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

View attachment 261815
Where ##O## is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where ##\vec F_i^S## is the specified force at a given point and ##\vec r_i## is the distance from ##O## to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\alpha} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\alpha} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\alpha$$

2) Determine the kinetic energy ##(T)## of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame ##XYS##. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it ##T_1##) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it ##T_2##).

- Let's determine ##T_1##: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine ##T_2##: Let's first get the coordinates of the center of mass of the ##AB## rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: ##\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y##.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the ##AB## rod, the squared modulus of the velocity (##|\vec v_{T_2}|^2##) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possess moment of inertia with respect to their center of mass (##\frac{mb^2}{12}## each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus ##T_2## is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy ##(T)## of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For ##x## we get

$$4m \ddot x = F$$

- For ##\theta## we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get ##a_c##

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$
Although, I couldn’t understand your solution (due to my lack of knowledge about Lagrangian Mechanics) but then also it was a beautiful solution, very well organised and clear.

JD_PM
etotheipi
2019 Award
My 2 cents worth contribution to this beautiful thread.

JD_PM
TSny
Homework Helper
Gold Member
Work-energy approach to problem 2

Discussions by @wrobel and others in the thread https://www.physicsforums.com/threads/acceleration-of-a-vertex.672835/ got me thinking about an energy approach to the problem.

The picture shows the applied forces in the center-of-mass reference frame. Each rod experiences an inertial force of magnitude F/4 as shown. The figure is shown for an arbitrary value of ##\theta##.

We will obtain the acceleration of joint C using the principle that the rate of change of kinetic energy of the system equals the rate at which the applied forces do work. The four inertial forces do zero net work in the center-of-mass frame since the horizontal displacements of the centers of rods AB and AD are to the right while the horizontal displacements of the other two rods are the same amount to the left. So, we only need to consider the work done by the applied force F at joint A.

It is not hard to show that the total kinetic energy of the system in the center-of-mass frame is ##T = \frac{2}{3}mL^2 \dot \theta^2##, where ##m## and ##L## are the mass and length of one rod. Also, the position of joint A is ##x_A = L \cos \theta##. The rate of change of ##T## equals the rate that ##F## does work:

##\dot T = F \dot x_A##

So, ##\frac{4}{3}mL^2 \dot \theta \ddot \theta = F\left(-L\sin \theta \, \dot \theta \right)##

Thus, ##\ddot \theta = - \large \frac{3F}{4mL}\sin \theta \,\,\,\,## [“pendulum” equation of motion for ##\theta##].

The position of joint C is ##x_C = -L \cos \theta##. Therefore,

##\ddot x_C = L \sin \theta \, \ddot\theta \,\,## whenever ##\dot \theta = 0##, as when ##F## is first applied. Substitute for ##\ddot \theta## to obtain

##\ddot x_C = -\frac{3F}{4m}\sin^2 \theta\,\,## when ##F## is first applied.

This gives the acceleration of joint C relative to the center-of-mass frame. Relative to the lab frame, the acceleration of the center of mass is ##\frac{F}{4m}## to the right. So, the acceleration of joint C relative to the lab is

##\ddot x_{C/ \rm lab} =\frac{F}{4m} - \frac{3F}{4m}\sin^2 \theta ##

or,

##\boxed{\ddot x_{C/ \rm lab} = \frac{F}{4m}\left(1-3 \sin^2 \theta \right)} \,\,\,## when ##F## is first applied.

For ##\theta = 45^o##, this gives ##\ddot x_{C/ \rm lab} = - \large \frac{F}{8m}##.

The value of ##\theta## for which ##\ddot x_{C/ \rm lab} = 0## when ##F## is applied is ##\theta = \arcsin (1/ \sqrt{3}) \approx 35^o##.

Delta2, JD_PM, mfb and 3 others
I vote for this solution by TSny ,at least this is only solution I understand

Delta2 and Spinnor
Chestermiller
Mentor
I vote for this solution by TSny ,at least this is only solution I understand
I'm amazed that you don't understand the solution I presented in post's #51 and 53; it is just straightforward freshman physics using force- and moment balances.

Last edited:
Spinnor
Gold Member
My 2 cents worth contribution to this beautiful thread.

Please note that I used the same approach that @wrobel at #21 and, to be fair, I checked it before getting the answer by myself.

I modified Chestermiller's nice diagram so that it can be applied to this approach

View attachment 261815
Where ##O## is at rest relative to the lab frame.

We first note that we have a time-dependent external force (exerted on A) which has no potential energy associated to it due to the fact that the force is not conservative.

At this point we think of the definition of generalized force (see for instance Classical Mechanics by Gregory, Chapter 12, page 337, definition 12.6)

$$Q_j = \sum_i \vec F_i^S \cdot \frac{\partial \vec r_i}{\partial q_j}$$

Where ##\vec F_i^S## is the specified force at a given point and ##\vec r_i## is the distance from ##O## to the given point.

Let's go step by step

1) Compute generalized forces

The generalized forces in this problem are:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial x}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial x}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial x}$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}+\vec F_B^S \cdot \frac{\partial \vec r_B}{\partial \theta}+\vec F_C^S \cdot \frac{\partial \vec r_C}{\partial \theta}+\vec F_D^S \cdot \frac{\partial \vec r_D}{\partial \theta}$$

Before computing anything we note that we've only been specified a force at A. Thus the rest are zero and we end up with:

$$Q_x = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial x}=F$$

$$Q_{\theta} = \vec F_A^S \cdot \frac{\partial \vec r_A}{\partial \theta}=-Fb \sin\theta$$

2) Determine the kinetic energy ##(T)## of the system

Note that we are going to work with two frames here: the lab frame (external observer) and the rhombic frame ##XYS##. König's theorem asserts that the kinetic energy of a system is the sum of two terms: the kinetic energy associated to the movement of the center of mass with respect to the lab frame (let me label it ##T_1##) and the kinetic energy associated to the movement of particles making up the system with respect of the center of mass of the system (let me label it ##T_2##).

- Let's determine ##T_1##: the center of mass (wrt the lab frame) is located at a distance OS, which is precisely our definition for the general coordinate x. Thus we get

$$T_1 = \frac 1 2 (4m) \dot x^2$$

- Let's determine ##T_2##: Let's first get the coordinates of the center of mass of the ##AB## rod

$$\Big( \frac b 2 \cos \theta, \ \frac b 2 \sin \theta \Big)$$

Then the velocity coordinates are

$$\Big( -\frac b 2 \sin \theta \dot \theta, \ \frac b 2 \cos \theta \dot \theta \Big)$$

Or in the more usual form: ##\vec v_{T_2}=-\frac b 2 \sin \theta \dot \theta \hat x + \frac b 2 \cos \theta \dot \theta \hat y##.

The nice thing here is that, although neither the coordinates nor the velocity coordinates of the other 3 rod's center of masses are the same as the ones for the ##AB## rod, the squared modulus of the velocity (##|\vec v_{T_2}|^2##) is.

$$|\vec v_{T_2}|^2=\frac{b^2}{4} \dot \theta^2$$

Besides, the rods possess moment of inertia with respect to their center of mass (##\frac{mb^2}{12}## each), which means that we have to take into account the kinetic energy associated to the moment of inertia of all 4 rods.

Thus ##T_2## is given by

$$T_2=4 \Big( \frac 1 2 m \frac{b^2}{4} \dot \theta^2 + \frac{1}{24} mb^2 \dot \theta^2 \Big)=\frac{2}{3} mb^2 \dot \theta^2$$

Thus the kinetic energy ##(T)## of the system is

$$T = 2m \dot x^2 + \frac{2}{3} mb^2 \dot \theta^2$$

3) Compute the equations of motion related to each generalized coordinate

The Lagrange equations for a general standard system are:

$$\frac{d}{dt} \Big( \frac{\partial T}{\partial \dot q_j} \Big) - \frac{\partial T}{\partial q_j} = Q_j$$

- For ##x## we get

$$4m \ddot x = F$$

- For ##\theta## we get

$$\frac 4 3 mb \ddot \theta = -F \sin \theta$$

4) Get ##a_c##

Based on the diagram we see that

$$a_c=\frac{d^2}{dt^2} OC, \ \ \ \ \theta=\frac{\pi}{4},\ \ \ \ OC=x-b \cos\theta$$

Thus, using the equations of motion we've got, we end up with

$$a_c= \ddot x + b\cos \theta \ddot \theta^2 = \frac{F}{4m}-\frac{3F}{4m} \sin \theta \cos \theta=-\frac{F}{8m}$$
You get the same answer for theta = 0 and 90 degrees, that result does not agree with my experiment but the results of post #58 do?

TSny
Homework Helper
Gold Member
I vote for this solution by TSny ,at least this is only solution I understand
Thank you, but both of @Chestermiller's solutions look correct to me.

Yes you are right