Moderate -0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions and finding the resultant. While it involves multiple forces at various angles (requiring trigonometry), it's a routine textbook exercise with a well-established method. The 6 marks reflect computational work rather than conceptual difficulty, making it slightly easier than average for A-level.
\includegraphics{figure_2}
Coplanar forces of magnitudes 16 N, 12 N, 24 N and 8 N act at a point in the directions shown in the diagram.
Find the magnitude and direction of the single additional force acting at the same point which will produce equilibrium. [6]
sign errors. Do not allow ‘forces to the left = forces to the right’
e.g. 12cos30−8sin30=16sin30 unless subsequently ‘corrected’.
(12sin30+24+8cos30−16cos30)=F or Fcos or Fsin
Answer
Marks
Guidance
x
A1
F = ( 30−4 3 ) =23.07 .
x
(12cos30−8sin30−16sin30)=F or Fsin or Fcos
Answer
Marks
Guidance
y
A1
F = ( 6 3−12 ) =−1.607 .
y
( )2 ( )2
F= 6 3−12 + 30−4 3
30−4 3
F =
cos(their )
6 3−12
F =
Answer
Marks
Guidance
sin(their )
M1
Attempt to find F.
Must have correct number of relevant terms. (Forces must have or
not have components as required). All forces resolved/not resolved
as appropriate, but allow consistent sin/cos muddle.
Allow use of their provided correctly derived from equations
with the correct number of relevant terms.
6 3−12
=tan−1
30−4 3
30−4 3
=cos−1 Note: this will not give the correct answer
their F
unless F given to several significant figures
6 3−12
=sin−1
their F
Answer
Marks
Guidance
M1
Attempt to find .
Must have correct number of relevant terms. (Forces must have or
not have components as required). All forces resolved/not resolved
as appropriate, but allow consistent sin/cos muddle. Allow upside
30−4 3
down so tan−1 .
6 3−12
Allow use of their F provided correctly derived from equations
with the correct number of relevant terms.
6 3−12 1.607
Note: watch for use of sin−1 or sin−1 which
30−4 3 23.07
leads to correct answer of angle 4.0° scores M0A0.
Answer
Marks
Guidance
Question
Answer
Marks
2
F = 23.1 N
=3.99 above the negative x-axis oe
A1
[23.1277…] Both correct Allow 4.0° but not simply 4° .
3.986... Allow answers about the direction such as ‘Above the
west’, ‘north of west’ etc, or clockwise 183.99 from x axis, or
resultant sketch with angle indicated. If not specified in working
please check original diagram to see if direction specified there
instead. Allow a bearing of 274.0°. Allow explanation of direction
that could be drawn uniquely.
Or e.g. 86.0° to left of the y-axis or 176.0° from the positive x-
axis.
6
Answer
Marks
Guidance
Question
Answer
Marks
Question 2:
2 | Resolving either direction | M1 | With correct number of relevant terms. Allow sin/cos mix. Allow
sign errors. Do not allow ‘forces to the left = forces to the right’
e.g. 12cos30−8sin30=16sin30 unless subsequently ‘corrected’.
(12sin30+24+8cos30−16cos30)=F or Fcos or Fsin
x | A1 | F = ( 30−4 3 ) =23.07 .
x
(12cos30−8sin30−16sin30)=F or Fsin or Fcos
y | A1 | F = ( 6 3−12 ) =−1.607 .
y
( )2 ( )2
F= 6 3−12 + 30−4 3
30−4 3
F =
cos(their )
6 3−12
F =
sin(their ) | M1 | Attempt to find F.
Must have correct number of relevant terms. (Forces must have or
not have components as required). All forces resolved/not resolved
as appropriate, but allow consistent sin/cos muddle.
Allow use of their provided correctly derived from equations
with the correct number of relevant terms.
6 3−12
=tan−1
30−4 3
30−4 3
=cos−1 Note: this will not give the correct answer
their F
unless F given to several significant figures
6 3−12
=sin−1
their F
| M1 | Attempt to find .
Must have correct number of relevant terms. (Forces must have or
not have components as required). All forces resolved/not resolved
as appropriate, but allow consistent sin/cos muddle. Allow upside
30−4 3
down so tan−1 .
6 3−12
Allow use of their F provided correctly derived from equations
with the correct number of relevant terms.
6 3−12 1.607
Note: watch for use of sin−1 or sin−1 which
30−4 3 23.07
leads to correct answer of angle 4.0° scores M0A0.
Question | Answer | Marks | Guidance
2 | F = 23.1 N
=3.99 above the negative x-axis oe | A1 | [23.1277…] Both correct Allow 4.0° but not simply 4° .
3.986... Allow answers about the direction such as ‘Above the
west’, ‘north of west’ etc, or clockwise 183.99 from x axis, or
resultant sketch with angle indicated. If not specified in working
please check original diagram to see if direction specified there
instead. Allow a bearing of 274.0°. Allow explanation of direction
that could be drawn uniquely.
Or e.g. 86.0° to left of the y-axis or 176.0° from the positive x-
axis.
6
Question | Answer | Marks | Guidance
\includegraphics{figure_2}
Coplanar forces of magnitudes 16 N, 12 N, 24 N and 8 N act at a point in the directions shown in the diagram.
Find the magnitude and direction of the single additional force acting at the same point which will produce equilibrium. [6]
\hfill \mbox{\textit{CAIE M1 2024 Q2 [6]}}