B Torsion Involving an Off Axis Thrust

[Devin-M]

I have tried to draw a force diagram to illustrate the problem. Suppose I have a 2m, 100kg uniform rod with wheels at the bottom. Center of mass is middle of rod. I want to accelerate this rod horizontally while it remains perfectly vertical (constant angle). But the force is coming from a rod on a car, and this rod is pushing horizontally 0.5m below the center of mass of the rod. To keep the 2m rod vertical, the moment thrust is applied I attach a 1 meter rod projecting horizontally from the front of the 2 meter rod, this rod is of negligible mass but has a weight at the end. How would I calculate the mass of the weight at the end for 1m/s^2 constant horizontal acceleration at constant angle vs 2m/s^2 constant horizontal acceleration at constant angle? Am I missing any variables?

 

[Baluncore]

You mark the centre of mass of the rod, but do not specify the mass of the wheels at the bottom of the rod. That will move the CofM of the vertical rod.

Is the vertical rod attached to the car, or only pushed?

 

[Devin-M]

Wheels negligible mass, only pushed.

 

[Baluncore]

The push force applied by the car will accelerate the total mass linearly forward, without rotation.
F = mt * a ; where mt is the total mass, including the forward mass.

That force will cause a torque F * 0.5 clockwise.
You need a front counter-weight, mc, to provide an F/2 counter torque.
That comes from; F/2 = mc * g * 0.5
mc * g * 0.5 = F/2
mc * g = F
mc = F / g
mc = a * mt / g

 

[berkeman]

I have tried to draw a force diagram to illustrate the problem.

Please make it a habit to always post your Free Body Diagram(s) (FBD) for such problems. Identify all of the forces on the body, and show your work to write the equations to sum all forces and torques.

That is how you should approach such questions and problems.

 

[Devin-M]

Is my explanation correct:

Since the counterweight is twice the distance from center of mass as the point of application of the force from the car, the force from the counterweight should be half the force from the car to prevent rotation.

Running some numbers, I get:

-with a force from car of 100 newtons, the counterweight mass is 5.1kg, total mass 105.1kg, acceleration .951m/s^2.

-with a force from car of 200 newtons, the counterweight mass is 10.2kg, total mass 110.2kg, acceleration 1.81m/s^2

 

[berkeman]

Is my explanation correct:

Since the counterweight is twice the distance from center of mass as the point of application of the force from the car, the force from the counterweight should be half the force from the car to prevent rotation.

Please make it a habit to always post your Free Body Diagram(s) (FBD) for such problems. Identify all of the forces on the body, and show your work to write the equations to sum all forces and torques.

Which word did you not understand?

 

[Devin-M]

Here are the equations and forces I used:

(100 Newton/2)/9.8 = 5.1 Counterweight Mass KG

Counterweight Mass 5.1 KG + Rod Mass 100 KG = Total Mass 105.1kg

Acceleration 0.951 m/s^2 = 100N / 105.1KG

 

[berkeman]

Please humor me. I'm a visual learner...

 

[Devin-M]

I will draw a better diagram sometime next 24 hrs.

 

[berkeman]

If "better" means FBD, then thank you.

https://www.ae.msstate.edu/vlsm/non_con_force/non_concurrent_forces/examples/example3.htm

 

[Devin-M]

One point I'm confused by... say we want to target a specific acceleration, say 1m/s^2, I can't figure out how to do this. It seems to calculate the force for 1m/s^2 acceleration we need to know the final mass including counterweight, but to know the mass including the counterweight we need to know the force.

 

[berkeman]

One point I'm confused by... say we want to target a specific acceleration, say 1m/s^2, I can't figure out how to do this. It seems to calculate the force for 1m/s^2 acceleration we need to know the mass, but to know the mass including the counterweight we need to know the force.

FBD...

 

[jrmichler]

We are trying to convince you that the key to solving this problem is an FBD. We do not expect you to know how to do this, but we do expect you to at least try.

Hint 1: It is not always possible to see the path to a full and correct solution. In that case, just start by doing what you can do. Normally, once you get part way to the solution, you will be able to figure out how to get all the way. Or we will help you.

Hint 2: If you do not know the value of a particular force, at least get the force arrow on the FBD, and mark it with a letter indicating force unknown.

 

[Baluncore]

It seems to calculate the force for 1m/s^2 acceleration we need to know the final mass including counterweight, but to know the mass including the counterweight we need to know the force.

Yes, that is correct, but that is how you specified the problem. All I did was to solve for your question, then I realised there was a special case solution.

There is an obvious acceleration limit with the fixed torque-arm lengths specified. For accelerations greater than g, the counterweight must weigh more than the total, which is slightly impossible, like having zero mass wheels, or zero rolling friction.

If the counterweight mass, mc, was specified, the length of the forward arm could be computed for any acceleration, and the acceleration limit would disappear from the solution. It is easier to slide a counterweight along a rod, than it is to adjust the mass of the counterweight.

I too, would like to see your FBD.

 

[Drakkith]

I suspect this is related to your skateboarding thread. If so, note that the diagram above is simplified beyond usefulness for that particular problem. Notably, you are missing a 'foot' of non-zero length that can apply a counter-torque to keep everything upright.

 

[Devin-M]

I was going to draw it like this but I have some doubts, as it would appear the center of mass moves when the front counterweight is attached.

I suspect this is related to your skateboarding thread. If so, note that the diagram above is simplified beyond usefulness for that particular problem. Notably, you are missing a 'foot' of non-zero length that can apply a counter-torque to keep everything upright.

This is a very different problem from the skateboard thread. The physical arrangement is different and that separate problem had many bio-mechanical issues (including the one you mentioned) and even psychological factors preventing arrival at a definitive solution.

 

[berkeman]

I was going to draw it like this but I have some doubts

What do the diagonal 50Nm vectors in the upper left represent? Why are there two of them?

 

[Devin-M]

What do the diagonal 50Nm vectors in the upper left represent? Why are there two of them?

One is a 50 newton meter torque around the center of mass from the car, the other is a 50 newton meter torque in the opposite direction from the counterweight.

 

[berkeman]

One is a 50 newton meter torque around the center of mass from the car, the other is a 50 newton meter torque in the opposite direction from the counterweight.

Okay, that's not how a traditional FBD is labeled, but I understand that you are new to that concept. In your FBD you should normally just show the forces and where they act on the structure. Then you write an equation to sum the forces on the structure and a separate equation to sum the torques (moments) on the structure about some point. If the structure is not moving (translating or rotating), then both of those sum equations are equal to zero. If the structure is translating or rotating, the sum of those forces or moments can result in a net acceleration (linear or angular).

Can you draw a simplified FBD now that shows the forces on the structure? And can you then write those two equations?

https://en.wikipedia.org/wiki/Free_body_diagram

 

[Devin-M]

If the counterweight is 5.1kg, I don’t know whether this weight is being transmitted to the ground since if the car wasn’t pushing the the whole arrangement forward, it would be in the process of tipping over.

 

[berkeman]

If the counterweight is 5.1kg, I don’t know whether this weight is being transmitted to the ground since if the car wasn’t pushing the the whole arrangement forward, it would be in the process of tipping over.

If you aren't sure what the value of the force to use for one of the forces acting on the structure in the FBD, just use a variable to label that force vector. You could use the vertical vector $\vec N$ for the force acting on the structure's wheel pointing upward, for example.

 

[Devin-M]

Do I need to start by picking a counterweight mass, then determining how far in front of the main rod the center of mass is located?

 

[berkeman]

Use variable names for now. Once you have your linear and rotational equations written, you can say what your goal is and solve the equations (like do you want to accelerate the car to balance the counterweight, etc.)...

 

[Devin-M]

Sorry it took so long, been busy with work. Here's what I've got so far:

I also found this equation to help in determining L1:

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html

Also I can use Pythagorean theorem C=sqrt(a^2+b^2) to determine the length of L2.

For the torque from the car and counterweight, I believe I can use this formula:

 

[Drakkith]

How accurate do you want to be? With a torque of 50 NM without the weight, you only need about a 5.1 kg weight to counter that torque in this setup. For a 2 m/s^2 acceleration that doubles to 10.2 kg. Whatever else may happen when the center of mass moves, those numbers should be pretty close to what you need.

 

[Devin-M]

According to this page, the moment of inertia becomes a factor which so far had not been considered:

https://physics.stackexchange.com/questions/55631/motion-of-space-ship-when-thrust-is-off-center

It’s not exactly the same problem, but an excerpt is as follows:

“The impulse will generate a force, F, in the usual way and push the spaceship in the direction of the rocket. The force exerted on the sphere in that direction can be analysed into the tangent and the perpendicular to the surface of the sphere. If θ is the angle between the rocket direction and the normal to the sphere we have:

Tangent component: FT=Fsin(θ)

Normal component: FN=Fcos(θ).

The normal component is parallel to the radius of the sphere and passes through the centre (CM) and has no moment with respect to the centre. This component will push the sphere in the normal direction.

The tangent component has a moment with respect to the centre

M=FRsin(θ).

This component would rotate the sphere, should the axis of the sphere be pivoted, but it is not! However, due to the inertia of the mass of the sphere, it would be sufficient to give a pivotal leverage for the tangent force to rotate the sphere. The law of conservation of energy must be written, for a short time interval of application of the force, as it moves the spaceship by a displacement x, in the form

F.x=1/2mv^2+1/2Iω^2 <—here

The first term on the RHS is the kinetic energy due to the linear motion, and the second is the kinetic energy due to the rotational motion…”

 

[Baluncore]

According to this page, the moment of inertia becomes a factor which so far had not been considered:

Your definition of the problem defined that away. You defined the acceleration and force to be balanced, so there is NO rotation.

The counterweight was at the same height, as the centre of mass of the vertical rod. For horizontal translation, the two could be simply added.

The moment of inertia is only important if rotation occurs, when the height of the counterweight, and that of the CofG of the vertical rod, will differ and change.

 

[Devin-M]

Thank you.

 

[Devin-M]

If we attach a bracket of negligible mass as shown, we can eliminate angle D and L2 from the previous force diagram, but we still need to determine L1:

 

[jrmichler]

Still no FBD after 30 posts, so time to fix that. The vertical rod weighs 100 kg, and the wheels at the bottom have zero mass, so that rod is shown as a 100 kg mass at the rod center of mass (COM), which is 1 m above the ground. A mass M is at the same height as the COM of the rod, and 1 m to the left of the rod COM. The assembly is vertically supported by a vertical force at the bottom of the rod; that force is $(100 + M) * 9.8$ N. There is no horizontal force at the bottom of the rod because these are physics wheels - zero mass, zero friction.

The total mass is 100 + M, and the acceleration, $a$, is unknown, so the total acceleration force is $(100 + M) * a$. There is an equal and opposite force $(100 + M) * a$ from the car applied at point P. The FBD shows this:

Everything is in equilibrium. The total vertical force down is $(100 + M) * 9.8$ N, which is balanced by the vertical force up at the wheels. The total horizontal force due to the acceleration is balanced by the force from the car. The only unknown is the acceleration, which is calculated by taking a sum of moments about a convenient point. The two most convenient points are the COM of the 100 kg rod, and the point P where the force from the car is applied. Arbitrarily choosing to take moments about point P, the calculation is as follows:

The mass M has a moment $M * 9.8 m/s^2 * 1 m = 9.8M$ Nm. The acceleration force has a moment $(100 + M) * a * 0.5m = 0.5 * a * (100 + M)$ Nm in the opposite direction. So the final equation for a is $$9.8M - 0.5 * a * (100 + M) = 0$$
One equation with one unknown, solve for a. Alternatively, you can define the acceleration with the mass M unknown, and solve for M.

 

[Baluncore]

To make the model easy to compute:
1. Keep it vertical and rotation free, by balancing the torques.
2. Specify the mass of counterweight, C. Let it slide along and clamp to a rod, at position L1.
3. The mass of that rod can be balanced by the mass of the car push rod.
4. The car pushes on a point L2 below the CofM.
5. That keeps the CofM above the wheel.

 

[Devin-M]

Correct me if I’m wrong, but in this drawing the counterweight M is unnecessary because the force from the car is being applied directly through the center of mass of the 100kg rod. The problem I wanted to solve for is when the force is applied 0.5m beneath the center of mass of the 100kg, 2m long rod (before counterweight added). The counterweight’s purpose is to counter the backwards tilting torque caused by the force being applied below the center of mass.

 

[DaveC426913]

If we attach a bracket of negligible mass as shown, we can eliminate angle D and L2 from the previous force diagram, but we still need to determine L1:

View attachment 330239

The addition of that bracket has exactly zero effect on the problem or solution.

 

[Devin-M]

I will attempt to solve this with a 5kg counterweight.

We have:
Counterweight: 5kg
Rod: 100kg

Center of Mass Equation

((5*0)+(100*1))/(5+100) = 0.9523m

^Center of mass is 0.95m from Counterweight

Torque from counterweight:

5kg*9.8m/s^2*0.95m = 46.55Nm torque from counterweight

Force from car:

46.55Nm / 0.5m = 93.1 N force from car

Acceleration from force & mass:

93.1N / (100kg + 5kg) = 0.88m/s^2 acceleration

 

[jrmichler]

Correct me if I’m wrong

Look carefully at the dimensions in the FBD. The portion of the vertical rod that is above the COM is not shown because it has no mass, and therefore no effect on the solution. The wheels are massless, so the FBD has exactly two masses, both of which are shown as point masses:
1) The COM of the vertical rod.
and
2) The added mass out in front.

Since this problem is defined as translation only, with zero rotation, all masses are shown at their COM. If you were working on a different problem involving angular acceleration, you would need to show distributed masses as distributed masses. The vertical rod would be one such mass.

The secret to solving these types of problems is to put most of your effort into getting the FBD correct, and without any extraneous detail. The best FBD has only the information needed to solve the problem.

If you want to better understand FBD's, solve this one several different ways:
1) By sum of moments about point P, the point of application of the force from the car.
and
2) By sum of moments about the COM of the vertical rod.
and
3) By sum of moments about the mass M.
and
4) By sum of moments about the center of the wheels.

According to this page,

Looking for an equation online will, all too often, lead you to a wrong answer because there is no guarantee that the online equation is correct for your FBD. If the FBD is correct, then the correct equation is easily found.

 

[Devin-M]

The portion of the vertical rod that is above the COM is not shown because it has no mass, and therefore no effect on the solution.

That portion weighs 50kg-- the 2m uniform rod is 100kg, so the half above that point is 50kg.

 

[Devin-M]

Only the horizontal rod has negligible mass.

 

[jrmichler]

OOPS, my bad. Good catch. But, because this problem is translation only, the vertical rod is shown as a 100 kg point mass.

 

[Devin-M]

OOPS, my bad. Good catch. But, because this problem is translation only, the vertical rod is shown as a 100 kg point mass.

The rod in the problem is not a point mass, so depending where the force is applied (below or above the center of mass), there is either a backwards tilting or forwards tilting torque from the thrust from the car (in this case a backwards tilting torque from the thrust from the car, because the thrust is applied 0.5m below the center of mass of the vertical rod).

 

[berkeman]

The rod in the problem is not a point mass, so depending where the force is applied (below or above the center of mass), there is either a backwards tilting or forwards tilting torque from the thrust from the car (in this case a backwards tilting torque from the thrust from the car, because the thrust is applied 0.5m below the center of mass of the vertical rod).

I haven't been following your thread lately, but my impression all along was that you want to do the calculation for the situation where the vertical rod does not rotate, right? So there are no MOI considerations...

 

[Devin-M]

I haven't been following your thread lately, but my impression all along was that you want to do the calculation for the situation where the vertical rod does not rotate, right? So there are no MOI considerations...

The problem is that without the counterweight, the main rod would tend to tilt backwards (top towards car) as it is accelerated from the thrust from the car below the rod's center of mass (horizontal thrust applied below center of mass of main 100kg uniform rod). The counterweight is added to counter this torque resulting in net 0 torque composed of 2 equal torques in opposite directions, and an acceleration from the force from the car in the horizontal direction.

 

[Baluncore]

I keep thinking that it will all end in a tangled mess. As the car accelerates steadily things are OK, but when the car becomes torque limited at high RPM, the counterweight will fall forwards, and the car will run over the contraption, hence, the tangled mess.

 

[Devin-M]

I believe I was able to solve counterweight mass KG for a desired acceleration...

A = Acceleration m/s^2 = 1
B = Counterweight Mass kg =
C = Car Force Distance From COM meters = 0.5
D = Main Rod Mass kg = 100
G = Acceleration of Gravity m/s^2 = 9.8

B = (D*C^2*A+D^2*C*A)/(D*G-C^2*A-D*C*A)

B = (100*0.5^2*1+100^2*0.5*1)/(100*9.8-0.5^2*1-100*0.5*1)

5.4kg = (100*0.5^2*1+100^2*0.5*1)/(100*9.8-0.5^2*1-100*0.5*1)

^5.4kg counterweight mass is needed for 1m/s^2 acceleration

 

[Baluncore]

What happens when you set the required acceleration to 10.0 ?

 

[Devin-M]

Please disregard the previous formula I posted-- it's in error.

I went back to specifying the counterweight mass (without knowing acceleration) and discovered there is a certain point when you keep increasing the counterweight mass the acceleration starts going back down... for example 300kg counterweight has lower acceleration than 100kg counterweight.

 

[Devin-M]

90kg Counterweight:
Center of Mass
((90*0)+(100*1))/(90+100) = 0.5263m
^Center of mass is 0.5263m from Counterweight
Torque from counterweight:
90kg*9.8m/s^2*0.5263m = 464.19Nm torque from counterweight
Force from car:
464.19Nm / 0.5m = 928.38 N force from car
Acceleration from force & mass:
928.38N / (90kg + 100kg) = 4.8862m/s^2 acceleration

100kg Counterweight:
Center of Mass
((100*0)+(100*1))/(100+100) = 0.5m
^Center of mass is 0.5m from Counterweight
Torque from counterweight:
100kg*9.8m/s^2*0.5m = 490Nm torque from counterweight
Force from car:
490Nm / 0.5m = 980 N force from car
Acceleration from force & mass:
980N / (100kg + 100kg) = 4.9m/s^2 acceleration

110kg Counterweight:
Center of Mass
((110*0)+(100*1))/(110+100) = 0.476m
^Center of mass is 0.476m from Counterweight
Torque from counterweight:
110kg*9.8m/s^2*0.476m = 513.128Nm torque from counterweight
Force from car:
513.128Nm / 0.5m = 1026.25 N force from car
Acceleration from force & mass:
1026.25N / (110kg + 100kg) = 4.88m/s^2 acceleration

300kg Counterweight:
Center of Mass
((300*0)+(100*1))/(300+100) = 0.25m
^Center of mass is 0.25m from Counterweight
Torque from counterweight:
300kg*9.8m/s^2*0.25m = 735Nm torque from counterweight
Force from car:
735Nm / 0.5m = 1470 N force from car
Acceleration from force & mass:
1470N / (300kg + 100kg) = 3.67m/s^2 acceleration

 

[Devin-M]

It looks like acceleration peaks at 1/2 of Earth's gravitational acceleration, when the counterweight mass equals the main rod mass, then acceleration decreases with additional counterweight mass... which implies you can't calculate the counterweight mass from the desired acceleration because there's more than 1 solution, and the peak horizontal acceleration is 1/2 of Earth's 9.8m/s^2 vertical gravitational acceleration.

 

[Devin-M]

If there were no counterweight and stable acceleration were achieved by setting an initial forward lean angle, would there be any further acceleration to be gained by using a counterweight in addition to the forward lean angle?

 

[Devin-M]

If there were no counterweight and stable acceleration were achieved by setting an initial forward lean angle, would there be any further acceleration to be gained by using a counterweight in addition to the forward lean angle?

The reason I ask is some electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…