CAIE M1 2023 March — Question 5 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionMarch
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyStandard +0.3 This is a straightforward 3D equilibrium problem with symmetry. Part (a) involves resolving forces vertically and horizontally with symmetric struts at 45°, requiring basic trigonometry and simultaneous equations. Part (b) adds a horizontal force and asks when one strut force becomes zero—a standard extension requiring minimal additional insight. The problem is slightly easier than average due to the symmetric geometry and standard angles.
Spec3.04b Equilibrium: zero resultant moment and force
\includegraphics{figure_5} The diagram shows a block \(D\) of mass 100 kg supported by two sloping struts \(AD\) and \(BD\), each attached at an angle of \(45°\) to fixed points \(A\) and \(B\) respectively on a horizontal floor. The block is also held in place by a vertical rope \(CD\) attached to a fixed point \(C\) on a horizontal ceiling. The tension in the rope \(CD\) is 500 N and the block rests in equilibrium.
  1. Find the magnitude of the force in each of the struts \(AD\) and \(BD\). [3] A horizontal force of magnitude \(F\) N is applied to the block in a direction parallel to \(AB\).
  2. Find the value of \(F\) for which the magnitude of the force in the strut \(AD\) is zero. [3]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks Guidance
5(a)Attempt to resolve vertically M1
A B
missing.
500+Tcos45+Tcos45−100g=0
OR 500+T cos45+T cos45−100g=0 AND
A B
T (sin45)=T (sin45)
AnswerMarks Guidance
A BA1 Must have T =T =T .
A B
Allow if 500−2Tcos45−100g=0.
Allow 500−T cos45−T cos45−100g=0 AND
A B
T (sin45)=T (sin45).
A B
AnswerMarks Guidance
T = 354 NA1 500
Allow 250 2, .
2
Allow if 500−2Tcos45−100g=0 to obtain
T = −354 and then state magnitude is 354.
If T and T are different values then A0.
A B
Alternative Method 1 for Question 5(a): Resolving perpendicular to a strut
Resolve perpendicular to T or T
AnswerMarks Guidance
A BM1 3 terms; allow sign errors; allow g missing.
T (or T )+500cos45=100gcos45
AnswerMarks Guidance
A BA1 Allow T (or T )+100gcos45=500cos45.
A B
T =T =354
AnswerMarks Guidance
A BA1 500
Allow 250 2, .
2
AnswerMarks Guidance
QuestionAnswer Marks
5(a)Alternative Method 2 for Question 5(a): Using triangle of forces
Attempt Pythagoras on a right-angled triangle of forces or use of
trigonometry
 T 2 +T 2 =(100g−500)2 
 A B 
 100g−500 100g−500
OR sin45orcos45= or
 
 T T 
AnswerMarks Guidance
A BM1 4 terms; allow with T and T ; allow sign errors; allow g
A B
missing.
T2 +T2 =(100g−500)2
OR T 2 +T 2 =(100g−500)2 AND T (sin45)=T (sin45)
A B A B
T (or T ) T (or T )
OR sin45= A B OR cos45= A B
AnswerMarks Guidance
100g−500 100g−500A1 T (or T ) T (or T )
Allow sin45= A B OR cos45= A B .
500−100g 500−100g
T =T =354
AnswerMarks Guidance
A BA1 500
Allow 250 2, .
2
Alternative Method 3 for Question 5(a): Using Lami’s Theorem
AnswerMarks Guidance
Attempt at LamiM1 Allow with T and T ; allow sign errors; allow g missing.
A B
100g−500 T (or T )
= A B
AnswerMarks Guidance
sin90 sin135A1 500−100g T (or T )
Allow = A B .
sin90 sin135
100g−500 T (or T )
Allow = A B .
sin270 sin45
T =T =354
AnswerMarks Guidance
A BA1 500
Allow 250 2, .
2
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(b)Attempt to resolve vertically and horizontally M1
allow g missing.
Must have T =0.
A
T cos45+500−100g =0 and
B
F−T sin45=0
AnswerMarks Guidance
BA1 Allow −T cos45+500−100g=0 and
B
F+T sin45=0
B
OR T cos45+500−100g =0 and
B
F+T sin45=0
B
OR −T cos45+500−100g=0 and
B
F−T sin45=0.
B
For both equations correct.
AnswerMarks Guidance
F=500A1 awrt 500 to 3sf.
Alternative Method 1 for Question 5(b): Resolving perpendicular to T
B
Attempt to resolve perpendicular to T
AnswerMarks Guidance
BM1 3 terms; allow sign errors; allow g missing.
Must have T =0.
A
AnswerMarks Guidance
Fcos45+500cos45=100gcos45A1 Allow −Fcos45+500cos45=100gcos45.
F=500A1 awrt 500 to 3sf.
QuestionAnswer Marks
5(b)Alternative Method 2 for Question 5(b): Using Lami’s Theorem
Attempt at LamiM1 Allow sign errors.
100g−500 F  T 
= = B
 
AnswerMarks Guidance
sin135 sin135 sin90A1 500−100g F 100g−500 F
Allow = or = or
sin135 sin135 sin45 sin45
100g−500 F 100g−500 F
= or = .
sin45 sin225 sin225 sin45
AnswerMarks Guidance
F=500A1 awrt 500 to 3sf.
Alternative Method 3 for Question 5(b): Using triangle of forces
AnswerMarks Guidance
Attempt use of trigonometry on right angled triangleM1 Allow sign errors; allow g missing.
F
tan45=
AnswerMarks Guidance
100g−500A1 F
Allow tan45=
500−100g
AnswerMarks Guidance
F=500A1 awrt 500 to 3sf.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Attempt to resolve vertically | M1 | 4 terms; allow with T and T ; allow sign errors; allow g
A B
missing.
500+Tcos45+Tcos45−100g=0
OR 500+T cos45+T cos45−100g=0 AND
A B
T (sin45)=T (sin45)
A B | A1 | Must have T =T =T .
A B
Allow if 500−2Tcos45−100g=0.
Allow 500−T cos45−T cos45−100g=0 AND
A B
T (sin45)=T (sin45).
A B
T = 354 N | A1 | 500
Allow 250 2, .
2
Allow if 500−2Tcos45−100g=0 to obtain
T = −354 and then state magnitude is 354.
If T and T are different values then A0.
A B
Alternative Method 1 for Question 5(a): Resolving perpendicular to a strut
Resolve perpendicular to T or T
A B | M1 | 3 terms; allow sign errors; allow g missing.
T (or T )+500cos45=100gcos45
A B | A1 | Allow T (or T )+100gcos45=500cos45.
A B
T =T =354
A B | A1 | 500
Allow 250 2, .
2
Question | Answer | Marks | Guidance
5(a) | Alternative Method 2 for Question 5(a): Using triangle of forces
Attempt Pythagoras on a right-angled triangle of forces or use of
trigonometry
 T 2 +T 2 =(100g−500)2 
 A B 
 100g−500 100g−500
OR sin45orcos45= or
 
 T T 
A B | M1 | 4 terms; allow with T and T ; allow sign errors; allow g
A B
missing.
T2 +T2 =(100g−500)2
OR T 2 +T 2 =(100g−500)2 AND T (sin45)=T (sin45)
A B A B
T (or T ) T (or T )
OR sin45= A B OR cos45= A B
100g−500 100g−500 | A1 | T (or T ) T (or T )
Allow sin45= A B OR cos45= A B .
500−100g 500−100g
T =T =354
A B | A1 | 500
Allow 250 2, .
2
Alternative Method 3 for Question 5(a): Using Lami’s Theorem
Attempt at Lami | M1 | Allow with T and T ; allow sign errors; allow g missing.
A B
100g−500 T (or T )
= A B
sin90 sin135 | A1 | 500−100g T (or T )
Allow = A B .
sin90 sin135
100g−500 T (or T )
Allow = A B .
sin270 sin45
T =T =354
A B | A1 | 500
Allow 250 2, .
2
3
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | Attempt to resolve vertically and horizontally | M1 | 3 terms vertically and 2 terms horizontally; allow sign errors;
allow g missing.
Must have T =0.
A
T cos45+500−100g =0 and
B
F−T sin45=0
B | A1 | Allow −T cos45+500−100g=0 and
B
F+T sin45=0
B
OR T cos45+500−100g =0 and
B
F+T sin45=0
B
OR −T cos45+500−100g=0 and
B
F−T sin45=0.
B
For both equations correct.
F=500 | A1 | awrt 500 to 3sf.
Alternative Method 1 for Question 5(b): Resolving perpendicular to T
B
Attempt to resolve perpendicular to T
B | M1 | 3 terms; allow sign errors; allow g missing.
Must have T =0.
A
Fcos45+500cos45=100gcos45 | A1 | Allow −Fcos45+500cos45=100gcos45.
F=500 | A1 | awrt 500 to 3sf.
Question | Answer | Marks | Guidance
5(b) | Alternative Method 2 for Question 5(b): Using Lami’s Theorem
Attempt at Lami | M1 | Allow sign errors.
100g−500 F  T 
= = B
 
sin135 sin135 sin90 | A1 | 500−100g F 100g−500 F
Allow = or = or
sin135 sin135 sin45 sin45
100g−500 F 100g−500 F
= or = .
sin45 sin225 sin225 sin45
F=500 | A1 | awrt 500 to 3sf.
Alternative Method 3 for Question 5(b): Using triangle of forces
Attempt use of trigonometry on right angled triangle | M1 | Allow sign errors; allow g missing.
F
tan45=
100g−500 | A1 | F
Allow tan45=
500−100g
F=500 | A1 | awrt 500 to 3sf.
3
Question | Answer | Marks | Guidance
\includegraphics{figure_5}

The diagram shows a block $D$ of mass 100 kg supported by two sloping struts $AD$ and $BD$, each attached at an angle of $45°$ to fixed points $A$ and $B$ respectively on a horizontal floor. The block is also held in place by a vertical rope $CD$ attached to a fixed point $C$ on a horizontal ceiling. The tension in the rope $CD$ is 500 N and the block rests in equilibrium.

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the force in each of the struts $AD$ and $BD$. [3]

A horizontal force of magnitude $F$ N is applied to the block in a direction parallel to $AB$.

\item Find the value of $F$ for which the magnitude of the force in the strut $AD$ is zero. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q5 [6]}}
This paper (7 questions)
View full paper
Q1 5 Q2 5 Q3 5 Q4 7 Q5 6 Q6 9 Q7 13