Which of these rods are in equilibrium?

[ymnoklan]

Homework Statement

A rod with a square cross-section is imagined to be supported by two smooth cylinders in four different ways (see figure). Draw all the forces acting on the rod in each case, and use what you know about equilibrium and torque to assess whether it is possible to keep the rod supported in each of these four configurations.

Relevant Equations

Sigma F = 0, Sigma tau = 0, tau = F*r*sin(theta), G = m*g

(My attempt of drawing the forces is the figure with the arrows). Intuitively I would say that non of the rods are supported in any of these configurations, but I struggle to give a good explanation for why. I could guess that B would be able to keep up, however summing the forces, it seems just as unbalanced as A (which obviously can’t be supported). What have I misunderstood/overlooked?

 

[haruspex]

I could guess that B would be able to keep up, however summing the forces, it seems just as unbalanced as A

You are not given the magnitudes, so how can you sum them?

 

[ymnoklan]

True. How am I supposed to solve this?

 

[kuruman]

You are saying that (A) is obviously not balanced and I agree. Can you express mathematically why? This should give you a clue about how to deal with the less obvious cases.

 

[ymnoklan]

You are saying that (A) is obviously not balanced and I agree. Can you express mathematically why? This should give you a clue about how to deal with the less obvious cases.

 

[ymnoklan]

You are saying that (A) is obviously not balanced and I agree. Can you express mathematically why? This should give you a clue about how to deal with the less obvious cases.

In B x>d, which indicates that the rod is supported

 

[kuruman]

In B x>d, which indicates that the rod is supported

Good. What about the other two cases?

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[ymnoklan]

Good. What about the other two cases?

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I don't think neither C nor D are supported as only the cylinder below the rod can exert an upward normal force. This cannot balance both force and torque.
Also, thank you for the feedback. I will do my best.

 

[kuruman]

I don't think neither C nor D are supported as only the cylinder below the rod can exert an upward normal force. This cannot balance both force and torque.

To guide your thinking (and the math) for (C) and (D) imagine that the top cylinder is a coin, the rod is a ruler and the bottom cylinder is your extended index finger. Your task is to balance the ruler + coin on your finger like a see-saw. Which of the two choices (if any) is more likely to work? Then put in the math to find the conditions for success or failure.

 

[haruspex]

I don't think neither C nor D are supported as only the cylinder below the rod can exert an upward normal force.

If it is not balanced, there must be a way the rod's mass centre can move downwards without either cylinder moving. In D, if the mass centre moves down then either the left end of the rod must move up or everywhere left of the mass centre must move down.

This cannot balance both force and torque.

It is not necessary for the cylinder below to do everything. The cylinder above can help. We only need that the net force from the cylinders is upwards, balancing the weight of the rod, and the net torque from the rods about the mass centre is zero.

 

[ymnoklan]

To guide your thinking (and the math) for (C) and (D) imagine that the top cylinder is a coin, the rod is a ruler and the bottom cylinder is your extended index finger. Your task is to balance the ruler + coin on your finger like a see-saw. Which of the two choices (if any) is more likely to work? Then put in the math to find the conditions for success or failure.

Thank you for the hint! Well, D is obviously a more likely candidate to work as the top cylinder changes the center of mass of the system towards the left. I really struggle with the math though.

 

[ymnoklan]

If it is not balanced, there must be a way the rod's mass centre can move downwards without either cylinder moving. In D, if the mass centre moves down then either the left end of the rod must move up or everywhere left of the mass centre must move down.

It is not necessary for the cylinder below to do everything. The cylinder above can help. We only need that the net force from the cylinders is upwards, balancing the weight of the rod, and the net torque from the rods about the mass centre is zero.

Here is my attempt at the math

 

[kuruman]

Your answers in (C) and (D) can be made mathematically simpler if you choose a different point about which you calculate torques. Note that if the sum of the torques is zero about one point, it is zero about all points. The rod is not going to suddenly accelerate if you change the reference point for the torques. So you might as well choose a convenient point which is usually the point about which the system will rotate if it is NOT in equilibrium. When the center of mass (CM) is directly above the reference point, the system will be in equilibrium in cases where gravity is the only external force acting.

The CM of the (top cylinder + rod) system is at a point between the midpoint of the rod and the top cylinder. Thus, the picture in (C) cannot work while the picture in (D) can.

In (C) and (D) the convenient reference point for torques, i.e. the pivot, is the cylinder below the rod. You might wish to redo the math with the reference point at the pivot to see how this makes the mathematical explanation easier and to find the equilibrium condition for (D).

 

[ymnoklan]

Your answers in (C) and (D) can be made mathematically simpler if you choose a different point about which you calculate torques. Note that if the sum of the torques is zero about one point, it is zero about all points. The rod is not going to suddenly accelerate if you change the reference point for the torques. So you might as well choose a convenient point which is usually the point about which the system will rotate if it is NOT in equilibrium. When the center of mass (CM) is directly above the reference point, the system will be in equilibrium in cases where gravity is the only external force acting.

The CM of the (top cylinder + rod) system is at a point between the midpoint of the rod and the top cylinder. Thus, the picture in (C) cannot work while the picture in (D) can.

In (C) and (D) the convenient reference point for torques, i.e. the pivot, is the cylinder below the rod. You might wish to redo the math with the reference point at the pivot to see how this makes the mathematical explanation easier and to find the equilibrium condition for (D).

Hm, I thought had chosen the bottom cylinder as my pivot point in configuration C already. In D, I did chose the top cylinder as pivot point indeed, but when I try to switch it, the math doesn’t look any simpler. Instead I get a situation where I can’t tell whether d1 or d2 is larger (or equal in size). Well, I guess the equilibrium condition for (D) is d2>d1, or do you mean something else? What have I done wrong?

 

[Steve4Physics]

Maybe simple, qualitative (without any maths) answers would suffice.

For example, for part A), here is a possible answer:

$F_A$ and $W$ produce clockwise torques about B. $F_B$ produces zero torque about B. Therefore there will be a net clockwise torque about B. Consequently the bar will rotate about B and fall. I.e. the rod isn't supported.

 

[kuruman]

The distance between cylinder 1 and the center of mass is irrelevant for the purposes of calculating torques. You need to find what mathematical condition must be satisfied for the distance between the CM and the pivot to be zero. The relevant distances are the lever arms, i.e. the distances $d_1$ and $d_2$ from the pivot to the point of application of the force of gravity. See drawings below for (C) and (D). Then setting the net torque equal to zero, should give you an expression that you can solve to find the needed mathematical condition.

 

[haruspex]

Here is my attempt at the math

Your diagram for D in post #12 has the mass centre between the two rods. This does not match the given diagram in post #1.

Maybe simple, qualitative (without any maths) answers would suffice.

I'm sure that was the intent of the question, but I suggested doing the algebra because @ymnoklan couldn’t see how balance would be achieved in any of the cases.