Developing an Intuition for Statics Problems

[lightlightsup]

Homework Statement

See image for question from: Fundamentals of Physics - 10E - Page 346 - Chapter 12 - Problem 21 (YouTube: y2u.be/ZY3-ZSQP1Mo).

Relevant Equations

For Part A, I was able to figure out the τ components around the hinge and solve for T. The τ parts are mostly intuitive for me in these problems.

Parts B and C confused me (though I solved them with YouTube help (see above)).

I'm having trouble developing intuition around ΣF=ma=0 in statics. Specifically, with regards to Normal or Reaction Forces.

My explanation for Parts B and C right now is:
Since the strut is not rotating (Στ(strut's bottom) = 0), all those forces on it will be directed down the strut and towards the hinge.
In this case, the downward Fg(block), Fg(strut), and the downward component of the T are all pushing down on the strut which is pushing into the hinge. The hinge responds with an equal force in the opposite direction (upward). So, we have no translational motion in the y direction.
The horizontal (-x, left) component of the T travels down the strut and tries to pull the hinge leftward (-x), but, once again, the hinge responds with an equal force in the opposite direction (rightward, +x) and so, we have no translational motion in the x direction.

Is this correct?
Does anyone have a more intuitive explanation or something I could watch or read to develop my intuition here? I know that not everything in physics has intuitive explanations but it's worth a try.

It just baffles me that I don't have to consider how far the block is from the hinge/strut connection when doing ΣF=0.

I know the general principles here are that:
ΣF(anywhere) = 0
Στ(anywhere) = 0

So, in a statics problem, since Στ(somewhere) = 0, then, I've got to trust that all those would be τs are being directed purely into ΣF=ma? I don't have to care about the distance from the point being considered during ΣF=ma? All the τs that would have gone into changing L are going to be re-directed into ΣF=ma (=0 due to a reaction force)?

 

[kuruman]

I'm having trouble developing intuition around ΣF=ma=0 in statics.

If a system is and remains at rest, the acceleration of its center of mass is zero. What is there to intuit?

My explanation for Parts B and C right now is:
Since the strut is not rotating (Στ(strut's bottom) = 0), all that force has to be re-directed somewhere.

All what force? The net force on the CM (we established) is zero. The equation $\Sigma \tau = 0$ says that the angular acceleration about the system's CM is zero. So we have two things happening simultaneously (a) the translational state of the CM is not changing and (b) the rotational state about the CM is not changing.

In this case, the downward Fg(block), Fg(strut), and the downward component of the T are all pushing down on the strut which is pushing into the hinge. The hinge responds with an equal force in the opposite direction (upward). So, we have no translational motion in the y direction.
The horizontal (-x, left) component of the T travels down the strut and tries to pull the hinge leftward (-x), but, once again, the hinge responds with an equal force in the opposite direction (rightward, +x) and so, we have no translational motion in the x direction.

Is this correct?

The best way to think about it is that all the forces involved (except gravity) will adjust themselves to provide zero linear and angular acceleration. That's what they will do if you move the hanging mass to a different position on the strut or the point where the cable is attached on the ground. Don't worry about where the forces "go". They don't go anywhere. Being contact forces, they adjust themselves to provide the observed accelerations which are zero.

 

[lightlightsup]

The best way to think about it is that all the forces involved (except gravity) will adjust themselves to provide zero linear and angular acceleration. That's what they will do if you move the hanging mass to a different position on the strut or the point where the cable is attached on the ground. Don't worry about where the forces "go". They don't go anywhere. Being contact forces, they adjust themselves to provide the observed accelerations which are zero.

That helps.

I'm trying to dumb down these problems into a simple algorithm I can use to solve them.
So far:
(1) establish that Στ=0 and resolve that equation.
(2) assume that all forces will now act translationally on the system.
(3) make sure that all the translational forces are placed correctly/logically.
(4) resolve the ΣFx/y=0 equations to ensure that the system is not moving in the y or x direction.

 

[haruspex]

That helps.

I'm trying to dumb down these problems into a simple algorithm I can use to solve them.
So far:
(1) establish that Στ=0 and resolve that equation.
(2) assume that all forces will now act translationally on the system.
(3) make sure that all the translational forces are placed correctly/logically.
(4) resolve the ΣFx/y=0 equations to ensure that the system is not moving in the y or x direction.

A step before that is to be clear how you are partitioning the system into subsystems and draw a free body diagram for each.
Next, identify the reactions between subsystems in contact. (A common blunder is to include forces from a subsystem that is at a second remove.). In this, consider whether there may be a torque in the reaction; if it is a free joint there will not be, but otherwise there might.
Then, for each subsystem pick your axis for torque analysis.
Now you are ready to write down the torque balance equation and pair of force balance equations for each subsystem.

Not sure what to say about your intuition that line of action of a force should still be relevant in the sum forces balance. Well, it isn't.
Indeed, counterintuitive situations do arise. If a force F is applied to a rod mass m floating in space then the linear acceleration is F/m regardless of where the force is applied on the rod. You may feel it should be less if applied at right angles to one end. The explanation is that, as a practical matter, it would be much more difficult for you, as provider of the force, to maintain that force when the rod offers little resistance. So it is hard to grasp intuitively applying the same force, on the one hand, in the middle of the rod or, on the other hand, at one end.

 

[lightlightsup]

free joint

By "free joint", you mean something that connects a system but cannot rotate? Like a rod or plank sticking out of a wall?

If a force F is applied to a rod mass m floating in space then the linear acceleration is F/m regardless of where the force is applied on the rod. You may feel it should be less if applied at right angles to one end. The explanation is that, as a practical matter, it would be much more difficult for you, as provider of the force, to maintain that force when the rod offers little resistance. So it is hard to grasp intuitively applying the same force, on the one hand, in the middle of the rod or, on the other hand, at one end.

See, this is the non-intuitive part for me.
Thanks for bringing this to my attention.
My understanding from this is that "how" the force is applied matters.
So, even if the force is continuously applied at 90° to one end, because there is no axis for the rod to rotate around: it will not rotate, it will only go straight in one direction?

 

[kuruman]

My understanding from this is that "how" the force is applied matters.
So, even if the force is continuously applied at 90° to one end, because there is no axis for the rod to rotate around: it will not rotate, it will only go straight in one direction?

Also important is where the force is applied because that determines the net torque on the extended body. In your rod example if you apply a force at some arbitrary point, the rod will acquire a linear acceleration of the center of mass $a_{cm}=F/m$ and an angular acceleration about the center of mass $\alpha =Fd/I$ where $d$ is the distance from the CM to the point of application of the force and $I$ is the moment of inertia about the CM. So to get the rod to accelerate without spinning, you have to app;y the force right at the CM ($d=0$). You can test this yourself. Put a meter stick flat on a smooth, bare floor, give it a sharp kick at various points and observe what it does immediately after the kick and before friction has any significant effect on its motion.

 

[lightlightsup]

The distance between the center of mass and the force (and angle) will determine how much of the force/work done will go into angular acceleration and how much will go into linear acceleration?

 

[kuruman]

I would not say force/work. The distance between the center of mass and the point of impact will determine how much angular momentum will be transferred, from zero (impact at the center of mass) to maximum (impact at the tip). The linear momentum transfer will be the same. The energy transfer depends on the nature of the impact, i.e. how much energy is lost to heat, sound, deformations, vibrations, etc.

 

[haruspex]

By "free joint", you mean something that connects a system but cannot rotate? Like a rod or plank sticking out of a wall?

No, quite the reverse. That would be a fixed joint. A free joint is one that can rotate freely, a frictionless hinge.

because there is no axis for the rod to rotate around: it will not rotate, it will only go straight in one direction?

Oh no, it will still rotate.
The force applied perpendicularly at one end will have a torque about the mass centre, so there will be an angular acceleration as well.

My point is that if the force F is maintained (let's say it is hooked to the end of the rod, so that as the rod rotates it is no longer perpendicular, but maintains the same direction) then there will be a linear acceleration F/m of the mass centre.

 

[lightlightsup]

Correct me if I'm wrong, but: When a sphere or other similar body rolls perfectly down an incline, the work done by $F_g$ gets converted into translational kinetic energy and rotational kinetic energy. How much of each depends on $I$. $K_R = \frac{1}{2}Iω^2$ and $K_T = \frac{1}{2}mv^2.$
Suppose I have a stationary rod in deep space and I momentarily apply a force at 90° to one end of the rod (a small flick, completely tangential). How much of this work will be converted into translational kinetic energy and rotational kinetic energy? Does this line of reasoning make sense?

 

[haruspex]

Suppose I have a stationary rod in deep space and I momentarily apply a force at 90° to one end of the rod (a small flick, completely tangential). How much of this work will be converted into translational kinetic energy and rotational kinetic energy? Does this line of reasoning make sense?

Apply an impulse p tangentially to one end of a stationary uniform rod, length 2L, mass m.
If the mass centre moves off at speed v we have p=mv.
If it rotates at rate ω, pL=Iω=ωmL²/3.
From these you can work out how much energy goes into each motion.

If the impulse p had been delivered to the centre of the rod, it would still have moved off at speed v, but not rotating, so less KE. This may seem surprising. To understand it, we need to think about how the impulse is delivered. If you hit the rod with a hammer at one end you will find it relatively difficult to impart much impulse. The same attack speed with the hammer at the middle of the rod will impart a greater p, so a greater v.

 

[lightlightsup]

So, back to the original problem:
My initial intuition was that some fraction of the forces at play were being "used up" to ensure that net torque remains at 0. It seems that I was wrong about the "used up" part.
It seems that once the net torque is locked in at 0, the translational forces come directly into play and in the problem given: they also cancel out in every axis and equal to 0.
For example: For the hinge to provide an upward normal force on the strut, it must first be locked in (net torque about it must equal to 0).

Take this mind twister:

L = length of uniform rod
m = mass of uniform rod
μs = coefficient of static friction
T = tension in cord
Fn = normal force from ground on rod
fs = static friction (directed right, +x)

Στ(bottom of rod)=0
+LTsinθ + - (L/2) mgsin(90-θ) = 0
T = (1/2)(mg) [sin(90-θ) / sin(θ)]

ΣFy,system=0
-mg + Fn = 0
Fn = mg

ΣFx,system=0
-T + fs = 0
-T + (μs)(Fn) = 0
(μs)(Fn) = T

Does this make sense?

 

[kuruman]

It does not make sense. Your last line (μs)(Fn) = T conflicts with what you got from the torque equation
T = (1/2)(mg) [sin(90-θ) / sin(θ)]. Just look at the drawing, the sum of the horizontal forces is zero which means that$f_s=T=\frac{mg}{2\tan\theta}~~~\left(\sin(90^{\text{o}-}\theta)=\cos\theta\right)$. The force of static friction is what is necessary to provide zero acceleration in the horizontal direction; $\mu_sF_n$ is the maximum value it can have. Also, this mind twister is a different problem and I don't see what it has to do with the original question.

Finally, I don't see what you mean by this

For the hinge to provide an upward normal force on the strut, it must first be locked in (net torque about it must equal to 0).

The hinge will provide an upward normal force whether the torque is zero or not. If you cut the cable and the strut accelerates down, the hinge will still provide an upward force. If it didn't, the strut would sink into the ground. You are looking for causal relationships where there aren't any.

 

[haruspex]

-T + (μs)(Fn) = 0

Only if we are told that the system is on the point of slipping. More generally, $|f_s|\leq\mu_s|F_n|$.
Other than that, your analysis looks right.

I wouldn't agree with your first "locking in" forces via torque balance view; you could equally think of locking in via linear force balance first. The equations operate simultaneously.

 

[lightlightsup]

The hinge will provide an upward normal force whether the torque is zero or not. If you cut the cable and the strut accelerates down, the hinge will still provide an upward force. If it didn't, the strut would sink into the ground. You are looking for causal relationships where there aren't any.

I wouldn't agree with your first "locking in" forces via torque balance view; you could equally think of locking in via linear force balance first. The equations operate simultaneously.

Yeah, I think I'm changing my conceptual model then.
The hinge will provide an upward force regardless is true.