A very simple moments question

[physicals]

Homework Statement

This is a past paper question that has appeared on the IAS level physics exam in the January series.

Relevant Equations

moments=fx

Hey there,
Before i introduce you to the question, i would like to clarify that this is not a home work question so i encourage all support that can be given without holding back, marking this as "cheating" (this has sadly happened to me in the past).

This question was part of the recent Edexcel IAS level physics unit 1 paper.
I had completed this paper myself, and its results were released just a few days ago. My marks (as always) has been exemplary but I am still skeptical about one question.

For some context, A level exam boards have strict mark schemes that deem to be a 100% correct. However, i believe, i have found a potential mistake in the mark scheme and would like some backup verification to my explanation.

Please refer to the question image below and its corresponding mark scheme explanation.
Though answer stated in the mark scheme is full correct, I disagree with the explanation given (which many students sadly loose marks for).

Here is my explanation:
P here is the thrust force in the diagonal rod
I believe that force F doesn't contribute to moments of its own since it must be balanced by a horizontal force in the opposite direction from the diagonal bar to maintain horizontal equilibrium. This means we can equate F=sin55*P. Now to find F we just need to find the value of P. We can construct another equation based on the principles of moments, 0.16*cos55*P=0.28*34, to find P. Using this we can easily find F.

The mark scheme, however, seems to ignore P and only focus on the "imagninary" moments due to F. I call this imaginary since the mark scheme seems to ignore the true moment effects of P is cancelled by the horizontal component of P.

Please share your thoughts on this problem.

.

 

[Steve4Physics]

For some context, A level exam boards have strict mark schemes that deem to be a 100% correct.

Pre-retirement I marked physics exam' papers for Edexcel (and for others).

Occasionally examination questions are found to be ambiguous/faulty. Similarly for mark schemes. It shouldn’t happen but it does.

Any problems are quickly discovered by the markers once the marking starts. Each problem is reported and guidance notices are then issued to the markers.

However, for the question you posted, there is nothing wrong with the mark scheme!

However, i believe, i have found a potential mistake in the mark scheme and would like some backup verification to my explanation.

Recent papers and mark-schems are only available to certain personnel (e.g. teachers at registered centres) - not to students. So it is not clear how you would have obtained the information.

P here is the thrust force in the diagonal rod

'P' is already used for a point on the diagram. It can cause great confusion to use the same symbol for different things. Please don't do it!

Let's use 'D' for the diagonal rod's compression. Then each end of the diagonal rod exerts an outwards force, of magnitude D, on whatever it is connected to.

I believe that force F doesn't contribute to moments of its own since it must be balanced by a horizontal force in the opposite direction from the diagonal bar to maintain horizontal equilibrium.

That's wrong. You can take moments about ANY point. If an object is in equilibrium the net moment will be zero (moments balance) whatever point you choose.

Choosing different points leads to different (correct) equations, possibly involving different forces.

We usually choose the point so that an unknown force passes through that point, therefore that force has a zero moment about the point. Then we can set-up a moments equation without worrying about that particular unknown force.

The reason the question instructs you to take moment about P is so that you can avoid having to find the compressive force (D) in the diagonal. You should always follow the instructions!

This means we can equate F=sin55*P. Now to find F we just need to find the value of P. We can construct another equation based on the principles of moments, 0.16*cos55*P=0.28*34, to find P. Using this we can easily find F.

You have come up with an alternative solution - but did not follow the instructions. If correct, you might be awarded 2 out of 4 marks.

The mark scheme, however, seems to ignore P and only focus on the "imagninary" moments due to F.

The moment of F about point P is very real!

I call this imaginary since the mark scheme seems to ignore the true moment effects of P is cancelled by the horizontal component of P.

Edit: Force P (what I call D) passes through point P. so it has zero moment about point P. The only forces (acting on the horizontal rod) which have moments about point P, are F and the 34N weight.

 

[Herman Trivilino]

I believe that force F doesn't contribute to moments of its own since it must be balanced by a horizontal force in the opposite direction from the diagonal bar to maintain horizontal equilibrium.

It's an equilibrium situation so all the forces are "balanced" by other forces. That's what it means to have a net force of zero.

Do you have any idea where your belief comes from? Learning involves examining your own cognitive processes.

 

[Lnewqban]

Here is my explanation:
P here is the thrust force in the diagonal rod
I believe that force F doesn't contribute to moments of its own since it must be balanced by a horizontal force in the opposite direction from the diagonal bar to maintain horizontal equilibrium.
...
The mark scheme, however, seems to ignore P and only focus on the "imagninary" moments due to F. I call this imaginary since the mark scheme seems to ignore the true moment effects of P is cancelled by the horizontal component of P.

The diagonal brace does not need to exist for the external forces and moment to remain the same.
We could visualize the represented armature as a L-shaped or T-shaped piece of steel, or as a solid polygon: the calculated values would be exactly the same.

In problems like this one, the analysis of the internal forces of the structure is irrelevant.
An input of force times lever combination induces a moment that travels through the armature and manifests itself as an output of force times lever combination the equal magnitude.

A force is pushing the wall horizontally at point P, having exactly the same magnitude as the force pulling the wall anchor horizontally at point F.
Simultaneously, 34 N of vertical force is loading the anchor, which is preventing the armature and the container from a free fall.

 

[haruspex]

force F doesn't contribute to moments of its own since it must be balanced by a horizontal force in the opposite direction from the diagonal bar to maintain horizontal equilibrium.

Just in case yet another way of putting it helps…
- A force cannot balance a moment. The moment of a force can balance a moment.
- Generally, the value of a moment is only meaningful if the axis is specified. (The exception is a pure torque, as arises as the result of two equal and opposite forces acting along different but parallel lines.)
If we take the axis as P, the compression in the support has no moment, but F does.
In your approach, you say that you found that support force using moments, but you do not say what axis you used.

 

[Lnewqban]

For the armature to remain in static equilibrium, external forces and/or moments must be balanced by opposite external forces and/or moments.
You will study what happens inside the armature at a later date.

 

[physicals]

Thank you all for your replies.

I must apologize for the mild ambiguity in my query. I do know that force F results in a moment about pivot P, but i meant that the clockwise moment due to F is exactly balanced by the anticlockwise moment due to the horizontal component of the thrust by the diagonal.

A force cannot balance a moment. The moment of a force can balance a moment.

This is essentially the basics of what i was trying to explain.

Anyways, asides from the "moments due to F" confusion, no one has acknowledged the thrust force due to the diagonal.

The reason the question instructs you to take moment about P is so that you can avoid having to find the compressive force (D) in the diagonal. You should always follow the instructions!

I agree that D has no effect at the pivot, but there must be force D from the diagonal onto the horizontal rod. This force is completely ignored by the mark scheme which is what i wanted to point out. This ignorance has also lead to a not so accurate explanation for the question.

Please consider the forces due to D.

 

[haruspex]

there must be force D from the diagonal onto the horizontal rod.

Yes.

This force is completely ignored by the mark scheme

because the hint says to take moments about P. Since the line of action of that force is through P, it has no moment about P. Hence we can safely ignore it.

 

[physicals]

Its starting to make sense now. But i still don't understand why we don't consider the physical effects of force D on the horizontal rod.

Maybe I should just accept that the mark scheme provides an alternate (more simpler) method for solving the question, and that both methods are equally plausible.

 

[kuruman]

Its starting to make sense now. But i still don't understand why we don't consider the physical effects of force D on the horizontal rod.

Perhaps a formal solution will clarify things for you.

Consider the free body diagram on the right. The system consists of the horizontal piece plus the diagonal piece. The vertical piece is not part of the system but is shown (faintly) inside the FBD for reference. There are three forces acting on the system.

1. Force $\mathbf F$ acting on the horizontal piece
2. Force $\mathbf D$ acting on the diagonal piece
3. The weight $\mathbf W$ acting on the horizontal piece.

These forces are resolved into vertical and horizontal components as shown. Because the system is in equilibrium,

1. The sum of the forces in the horizontal direction is zero
2. The sum of the forces in the vertical direction is zero
3. The sum of the moments about point P is zero.

Using standard conventions, the corresponding equations are $$ -F_x +D_x =0 \tag{1} $$ $$F_y+D_y-W =0 \tag{2} $$ $$ F_xh-WL_2 =0 \tag{3} $$ Equation (1) says that the normal force exerted by the vertical force piece on the diagonal piece is equal in magnitude and opposite in direction to the horizontal force exerted by the horizontal piece on the vertical piece. Although $F_x$ is what we are looking for, this equation cannot give us a numerical value.

Equation (2) says that the sum of the vertical forces exerted by the vertical piece must balance the suspended weight. Note that, one cannot separate how much of the weight is $F_y$ and how uch $D_y$. The contact at P could be frictionless ($D_y=0$) and the system will still be at equilibrium as long as $D_x\neq 0.$ Thus, all the diagonal piece has to do is ensure that the vertical piece provides a couple consisting of equal and opposite forces. The role of the diagonal force $\mathbf D$ is to adjust itself to provide the observed zero angular acceleration. Of course, being a contact force, it can only do that up to a certain limit. If you suspend a container ship instead of just a container, the system will collapse.

Equation (3) gives us the answer we are seeking with $h=\dfrac{L_1}{\tan(55^{\circ})}$ as shown in the solution.

 

[Herman Trivilino]

Its starting to make sense now. But i still don't understand why we don't consider the physical effects of force D on the horizontal rod.

Because in problems like this we assume the rod is rigid. That is, it won’t bend. Due to $\vec{D}$ or any other force.

 

[Steve4Physics]

Its starting to make sense now. But i still don't understand why we don't consider the physical effects of force D on the horizontal rod.

It might help to remember that there are two types of equilibrium – translational and rotational.

In this question, in the horizontal direction, translational equilibrium requires that $D \sin(55^\circ) = F$. Part of D’s ‘physical affect’ is to contribute to translational equilibrium.

Now consider rotational equilibrium. Moments about P must balance to avoid angular acceleration about P. The only forces with moments about P are F and the 34 N weight, This allows us to answer the problem without considering D.

Maybe I should just accept that the mark scheme provides an alternate (more simpler) method for solving the question, and that both methods are equally plausible.

If possible, try to understand why the different methods work. This will give you more options when problem-solving in the future.

 

[kuruman]

To complete the formal solution in post #10.

Suppose this problem had a part (b) asking you to find force $\mathbf D$ exerted by the diagonal piece on the horizontal bar. The equations in post #10 are of no help. Now what?

To answer part (b) we choose our system to be the horizontal bar only. The FBD looks the same except that the diagonal piece has been replaced by the force $\mathbf D$ that it exerts on the bar (see figure on the right.) We write the three equilibrium equations as before. However, with commendable foresight, we move point P (about which we calculate moments) to the left tip of the bar. This eliminates the need to calculate moments generated by $\mathbf F$.
$$ -F_x +D \sin\theta =0 \implies F_x=D \sin\theta \tag{1} $$ $$F_y+D\cos\theta-W =0 \implies F_y=W-D\cos\theta \tag{2} $$ $$ D\cos\theta~L_1-WL_2=0 \implies D=\frac{WL_2}{L_1\cos\theta}. \tag{3} $$ Equation (3) gives the magnitude of the diagonal force $D$ in terms of the given quantities. Replacing this value in the other two equations gives the components of $\mathbf F$,
$$F_x=W\left(\dfrac{L_2}{L_1}\right)\tan\theta~;~~F_y=W\left(1-\frac{L_2}{L_1}\right).$$ (Added on edit)
Do you see how it works? This solution finds all the forces there are, the "official" solution finds only what is being asked, namely $F_x.$

 

[physicals]

The pieces finally fit now! I understand that D is the true physical effect of the diagonal rod's force, we only choose to resolve the vector to make calculations easier. So my assumption that the vertical and horizontal components of D are the only "physical" force effects is wrong.

However, my method of solving is not wrong in any way. Its just an alternate method to the one given in the mark scheme!

 

[physicals]

My doubt has finally been clarified, but for the sake of physics debate and to ensure i have no knowledge gaps, i am not fully convinced by the direction of the forces on @kuruman diagram explanation.

Though i fully understand the logic explained by the diagram, i think force F(y) should be in the other direction, because the moments about any point must be 0 to maintain equilibrium, and taking moments about the point of contact with the diagonal rod and the horizontal rod, there would clearly be a net clockwise moment so the system wont be in equilibrium. Am i correct or am i missing something?

To answer part (b) we choose our system to be the horizontal bar only. The FBD looks the same except that the diagonal piece has been replaced by the force D that it exerts on the bar (see figure on the right.) We write the three equilibrium equations as before. However, with commendable foresight, we move point P (about

 

[anuttarasammyak]

All external forces should be considered at the points where they act on the body. D is acting on point Q in the figure you show. But D, as an external force, acts on P.

Momentum balance around P
$$L_2w=hF_x$$
Momentum balance around Q
$$(L_2-L_1)w+L_1F_y+L_1D_y=hD_x$$
They are equivalent thanks to balance of forces
$$F_y+D_y=w$$
$$F_x=D_x$$

 

[physicals]

I still think F(y) must be in the other direction

 

[anuttarasammyak]

$$F_y=w-D_y$$
it can be negative, zero or positive.
Some example cases
$$F_y=w > 0,\ \ D_y=0$$

$$F_y=0,\ \ D_y=w$$

$$F_y<0,\ \ D_y=w-F_y>w$$

 

[A.T.]

Though i fully understand the logic explained by the diagram, i think force F(y) should be in the other direction

The arrow in the diagram just indicates the positive direction for a force component (sign convention), not the actual force direction, which is often unknown when the diagram is drawn. Since the magnitude derived for Fy is negative for L1 < L2, Fy is indeed opposite to its arrow direction for the geometry shown in the diagram.

 

[Steve4Physics]

I still think F(y) must be in the other direction

To add to what’s already been said, you are correct but it doesn't matter.

If you calculate $F_y$ you will get a negative value. You will then know the actual direction is opposite to the initial ‘guess’.

This approach is valuable in complicated systems, where it may not be easy to find all force directions in advance.

We do a similar thing when analysing complex electrical circuits. If the direction of the current in part of a circuit isn't obvious, we guess. If the guess turns out wrong, we get a negative value.

 

[A.T.]

This approach is valuable in complicated systems, where it may not be easy to find all force directions in advance.

It's also useful for expressing general solutions, where a quantity can have either direction, depending on some free parameters.The system doesn't have to be complex.

 

[kuruman]

Though i fully understand the logic explained by the diagram, i think force F(y) should be in the other direction, because the moments about any point must be 0 to maintain equilibrium, and taking moments about the point of contact with the diagonal rod and the horizontal rod, there would clearly be a net clockwise moment so the system wont be in equilibrium. Am i correct or am i missing something?

You are correct but you are missing the final point enunciated by @A.T. in ##19 and @Steve4Physics in #20. This is summarized as "Derive algebraically once, calculate numerically as many times as you wish." To illustrate further, suppose that the container were suspended between the vertical piece and the point of contact with the diagonal piece as shown in the figure on the right.

Is it obvious in this case whether the net moment about your choice of reference point P is clockwise or counterclockwise? Clearly, $F_y<W$ but its point of application is farther from point P than the point of application of $W$. To make it obvious, you might consider moving point P yet again to the point where the weight is suspended. This doesn't help much as you can see.

All falls into place if you apply the following method

1. Choose positive directions for Cartesian axes $x$ and $y$ and for moments. Conventionally, $x$ is positive to the right, $y$ is positive from bottom to top and moments are positive out of the screen/page/counterclockwise.
2. Draw a free body diagram using arrows to represent forces. Label each arrow with a symbol representing the magnitude of the force.
3. Choose a convenient reference point for expressing moments, usually where more than one forces are applied.
4. Write the three equilibrium equations and find algebraic expressions for the unknown quantities.
5. Substitute numbers to find the unknown force(s). If the magnitude of a force turns out to be negative, then you have drawn its direction in the diagram incorrectly. You may correct the diagram if you so wish, but you have achieved your goal of finding the answer.

You see now that the equations derived in post #13 are applicable to the configuration in this post too. Since $F_y=W\left(1-\dfrac{L_2}{L_1}\right)$ and $L_2<L_1$, $F_y$ is positive which means that it was drawn correctly.

"Derive algebraically once, calculate numerically as many times as you wish."

 

[A.T.]

it can be negative, zero or positive.
Some example cases
Fy=w>0, Dy=0

Fy=0, Dy=w

In your examples the connections must transmit a torque.

In the solution given by @kuruman they are assumed to be torque free. Based on this assumption you can still make Fy positive by moving the load w closer to the wall, so L1 > L2, as shown below.

 

[Lnewqban]

... there would clearly be a net clockwise moment so the system wont be in equilibrium. Am i correct or am i missing something?

... but the system is in static equilibrium.
You are incorrect, because you are confusing a node or point of intersection of two members with a system.

By the way, point D is not said or represented to be a hinge, and therefore, it transfers a moment to the diagonal brace as well, which becomes another force to consider acting at point P, which subtracts from your Fy.

What do you consider to be your system in static equilibrium?

 

[anuttarasammyak]

In your examples the connections must transmit a torque.

Thanks. My thread examples were not appropriate. Net zero inhomogeneous x force act at the contact area of the remaining bar so that it produces counter torque, which should be treated by advanced mechanics.

 

[physicals]

Thank you all for clarifying my doubts! @A.T., especially, has given explanations with the most concise, understandable language without too much jargon (suits best for a high schooler).

Now that we have come to the end of the discussion, Im not sure how you end a tread, do you just stop replying, or do you click something to end it?

 

[physicals]

... but the system is in static equilibrium.
You are incorrect, because you are confusing a node or point of intersection of two members with a system.

By the way, point D is not said or represented to be a hinge, and therefore, it transfers a moment to the diagonal brace as well, which becomes another force to consider acting at point P, which subtracts from your Fy.

What do you consider to be your system in static equilibrium?

It just makes intuitive sense for f(y) to be in the other direction (or negative).

In the solution given by @kuruman they are assumed to be torque free. Based on thIs assumption you can still make Fy positive by moving the load w closer to the wall, so L1 > L2., as shown below.

Like @A.T has pointed out, f(y) would be positive or negative depending on the position of W. Since on the image W is to the right of the pivot formed due to the diagonal rod, f(y) must be negative.