Moments questions

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• Edexcel 9MA0 · June 2024 · Paper 32 · Q6 · 9 marks

  9 marks

  6

  
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  Examiner report

  The most successful candidates in this question tended to indicate what they were doing at each stage – taking moments about A, resolving horizontally, etc, in words or symbols. It is always good exam technique to annotate the diagram to support the answer. The majority of errors affecting performance in the question as a whole were due to incorrect force diagrams: the normal reaction force at C was sometimes drawn vertically upwards rather than perpendicular to the rod, and an extra reaction force was occasionally put in at G and/or B. Most candidates worked with the angle theta, but a few did convert into terms such as ( ) O sin 90  − and used these successfully. (a) There were a good number of fully correct responses to this ‘Show that’ question. The majority of candidates found and used the trig ratios successfully, possibly because the answer was given in the question, which allowed them to identify whether they had calculated or used the wrong one. If a candidate gained one or two marks, but not all three, it was usually due to sin/cos confusion or making arithmetic errors when rearranging the equation. Some candidates lost all the marks because they omitted the a’s and so did not have a moments equation, but most included these initially and then cancelled appropriately. The final mark was occasionally lost because explicit use of 3 cos 5 = was not seen or the candidate chose to give the force another name (such as RC) and so did not arrive at the printed answer. A common error was using 2a instead of 1.5a as the distance. A few candidates gave the impression of having arrived at the required answer, but on closer inspection were not correct in their work. The most common example of incorrect working was 2 0.5 sin 5 S Mg Mg  = = . Some tried to formulate a proof based on the given answer that did not involve taking moments, but instead tried to resolve forces, and received no credit. Most candidates presented their answer in the given form, although S on the RHS was occasionally seen. (b) Many candidates scored all six marks in this part. Resolving vertically and horizontally was the most common and the most successful method. There were few errors apart from sin and cos confusion. Taking moments about C to give an equation in R and F and then resolving was the next most common method, but candidates were more likely to make errors using this approach, either by confusing sin with cos or by omitting a force entirely. Very few candidates resolved parallel and perpendicular to the rod and most of those who used this method failed to arrive at the correct answer due to the more challenging algebra. Occasionally candidates wrote dimensionally inconsistent equations by failing to include lengths in all of their moment terms or, more rarely, erroneously including lengths in their force terms. Errors included missing a g where there should have been one or not having the correct number of terms in a moment or a resolution equation. A surprisingly common error was for students to confuse the reaction force S with the reaction force at A, resulting in the incorrect equation 2 5 F Mg  =  Some candidates lost method marks by resolving forces that did not need resolving (for example, by writing an equation with both sin Mg  and cos S ). Whichever method was used, early substitution for F into their solutions was often seen, with almost all candidates realising that they needed to use F = μR. Some candidates clearly still incorrectly believe that μ cannot be greater than 1 and were ‘dismissive’ when their value was. It was not uncommon to see very messy responses, with multiple attempts and work crossed out. Many candidates made a good attempt at solving for µ, but this was often impeded by the lack of clarity over the previous equations. The mark scheme enabled candidates who used a correct method but who made sign errors, confused sin and cos or used incorrect distances to score half marks (generally, M1A0M1A0M1A0), while those with more fundamental misunderstandings were prevented from accessing the dependent method mark; this seemed to sort out the candidates well. When µ was found, most candidates gave the answer as a fraction or to an appropriate degree of accuracy.

  Text version

  A uniform rod AB has mass M and length 2a. End A rests on rough horizontal ground. The rod is in equilibrium against a small smooth fixed horizontal peg P. The contact point on the rod is C, where AC = 1.5a. The rod makes angle theta with the ground, tan(theta) = 4/3. The peg's normal reaction on the rod at C is S.

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• Edexcel 9MA0 · June 2023 · Paper 32 · Q6 · 13 marks

  13 marks

  6

  
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  In the first part, the majority of candidates stated that the frictional force would act towards the right on the diagram and many gave a satisfactory reason. A minority gave the reason that the friction had to balance the reaction force at the wall which acted to the left. Most mentioned that it prevented the rod from slipping to the left. Those who did not achieve the mark often used language that implied the rod was actually moving. A small number thought that the direction was left or gave no reason at all. In part (b), the question directed the candidates to take moments about A and most were able to do so successfully with many achieving the three available marks. The occasional incorrect response had the distance to the reaction at B unresolved as 2a. Also a small number had a’s missing from both sides of their equation. Since this was a “show that” question it was important that all stages of the solution were shown and were correct. It should be noted that the candidate’s final answer had to be written exactly as the printed answer, using the same symbols in the same order, to achieve the final mark. Part (c) required a calculation of the coefficient of friction. There were many fully correct solutions involving vertical and horizontal resolution but many incorrect responses were also seen. The majority who tried other approaches, such as moments about B or resolving parallel and perpendicular to the rod, were usually unsuccessful as they would often miss out a term or there would be a sin/cos error. Some candidates attempted to resolve vertically and horizontally but included an extra force. It was not uncommon for the horizontal resolution F = S to be correct but an extra force would appear in the vertical resolution. Almost all recalled F = µR and tried to use it in some way. Part (d) involved finding the magnitude of the resultant force acting at A. It was not always well answered and sometimes omitted entirely. A significant minority of responses were, however, fully correct and there were many who, having made an error in part (c), were still able to apply Pythagoras’ theorem correctly with their values of F and S. However, some candidates did not understand what was required with R = mg, or an attempt to just combine R and F by adding, seen quite often. A small number failed to deal properly with the square root making an error with the simplification in terms of mg in an otherwise correct solution. In the final part, the majority of candidates were unable to gain the mark often because of lack of precision in the wording of their response. Many stated that S was larger because the weight was closer to B or it exerted more force/pressure at B. Some who did refer to an increased moment about A failed to specify it as the moment of the weight. Some mentioned distance but did not explicitly refer to moments. Candidates who used their moments equation to show that a distance of the weight from A greater than a led to a larger value of S were generally successful. A small number thought S was smaller or did not change.

  Text version

  A rod AB has mass M and length 2a. End A rests on rough horizontal ground and end B rests against a smooth vertical wall to the right of A. The rod makes an angle theta with the ground and is at rest in limiting equilibrium. The wall's normal reaction at B has magnitude S and acts horizontally to the left. In an initial model the rod is uniform, so its weight Mg acts at its midpoint.

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• Edexcel 9MA0 · June 2022 · Paper 32 · Q4 · 11 marks

  11 marks

  4

  
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  Examiner report

  In part (a), very few candidates gave a fully correct reason for why the frictional force must act to the right on the diagram. Many stated that friction must oppose the direction of motion, saying that the rod would otherwise slip to the left. Some did mention it opposing the horizontal component of the tension but then failed to include that friction was the only other horizontal force in the system which was a key part of the argument. In the second part, it was required to derive a given expression for T (tension). Although some candidates struggled to make a valid start and basically just wrote down the answer and stated ’by resolving ’, it was well done by a fair number who set up a correct equation for moments about A and hence deduced the given result. Alternative approaches by either taking moments about other points or resolving forces were very rare and almost never successful, mostly because of missing terms. A few candidates failed to use perpendicular distances for the moments of the weight terms or failed to include the weight of the rod leading to a missing term. Part (c) was generally well answered by those candidates who resolved vertically, substituted for T and re-arranged to obtain the given answer thereby gaining full marks in a few lines of working. Candidates who used alternative equations as a starting point were generally unsuccessful, usually a result of a missing term in one or both equations used, this often involving friction. In part (d) those candidates who resolved horizontally to give F = T sinƟ usually managed to score all four marks for deriving the exact value of the coefficient of friction. There were a variety of equations, resolving or taking moments, which would have led directly to a value for F since the tension and normal reaction were known at that stage; however, these involved more working and therefore greater opportunity for error. Nevertheless, successful solutions by these methods were seen on occasion. Most of those candidates who found an expression for F were able to achieve the independent method mark for the use of F = μR even if their method for finding F was incorrect. Since all the answers were given in this question, candidates needed to ensure that they showed sufficient detail in their working to warrant being awarded all the available marks. In particular, candidates should be encouraged to state their resolved forces and moments as equations, rather than considering them in isolation, and to ensure the substitution stages are shown.

  Text version

  A uniform rod AB has mass M and length 2a. A particle of mass 2M is attached at C, where AC = 1.5a. End A rests on rough horizontal ground and the rod is held in equilibrium at angle theta above the ground by a light string attached at B. The string is perpendicular to the rod and runs upwards and to the left from B. Its tension is T.

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• Edexcel 9MA0 · October 2021 · Paper 32 · Q3 · 10 marks

  10 marks

  3

  
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  Part (a) required candidates “ To show that…”. Consequently, full marks were only awarded if there was a fully correct solution, with all the steps included. Thus successful candidates needed one moments equation together with a vertical resolution and/or a horizontal resolution, depending on which point they had used for their moments equation. These equations, combined with F R µ ≤ , could then lead to full marks being earned. Many candidates assumed that the beam was in limiting equilibrium i.e. that F R µ = , produced an equation and then changed it to the required inequality. These candidates lost the final mark. Other errors included omission of g, sin/cos confusion in moments equations and assuming that the reaction at A and the reaction at B were the same. Candidates tended to be more successful in part (b), where a similar strategy was required, together with F R µ = , since the beam was now in limiting equilibrium. A common error was to have the friction force acting towards the wall. This lead to a k value of – 0.1 and the negative sign should have flagged up an error to candidates.

  Text version

  A beam AB has mass m and length 2a. It rests in equilibrium in a vertical plane perpendicular to a wall, with A on rough horizontal ground and B against a smooth vertical wall. The beam is inclined at angle theta to the horizontal and is modelled as a uniform rod. The coefficient of friction at A is mu.

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• Edexcel 9MA0 · October 2020 · Paper 32 · Q4 · 10 marks

  10 marks

  4

  
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  A significant number of candidates made little or no progress with this question; a number of responses showed a complete lack of understanding of the nature and directions of the forces acting on the ladder. In part (a) a fair number attempted to take moments about A to derive the given reaction at C although sometimes perpendicular distances were not used and this invalidated the method. The most straightforward approach for part (b) was to resolve horizontally and vertically. Some candidates chose to take moments about various points or to resolve perpendicular and/or parallel to the ladder. These methods often led to errors including missing terms and incorrect distances. Some thought the normal reaction at C was horizontal. Those who attempted to use a reaction perpendicular to the ladder generally used this in F=µR and were unaware that this is only valid for a normal reaction. Confusion between sin/cos in resolving the reaction at C horizontally and vertically was surprisingly common. There were, nevertheless, some well-expressed fully correct solutions seen.

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  A uniform ladder AB has mass M and length 6a. End A rests on rough horizontal ground. The ladder rests at point C against a fixed smooth horizontal rail, with the vertical plane containing AB perpendicular to the rail. Point C is 4a vertically above the ground. The ladder is inclined at angle alpha to the horizontal, where sin(alpha) = 4/5. The coefficient of friction at A is mu and the ladder is in limiting equilibrium. Since AC sin(alpha) = 4a, AC = 5a; the ladder's centre of mass is 3a from A.

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