Standard +0.3 This is a standard equilibrium on a slope problem requiring resolution of forces in two directions and application of friction F ≤ μR. The setup is routine for M1, though finding the 'least value' requires considering friction direction, adding a small problem-solving element beyond pure recall. The 6 marks reflect multiple steps but the techniques are standard textbook material, making it slightly easier than average.
\includegraphics{figure_6}
A particle of mass \(1.2\) kg is placed on a rough plane which is inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{3}{5}\). The particle is kept in equilibrium by a horizontal force of magnitude \(P\) N acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and the plane is \(0.15\).
Find the least possible value of \(P\). [6]
Attempt at resolving perpendicular to the plane to get an equation
*M1
mix, allow g missing.
For reference R=1.2gcos16.26...+Psin16.26... - allow with
an angle of 16 or better.
24 7
R=1.2g +P
Answer
Marks
Guidance
25 25
A1
288 7
R=11.52+0.28P or R= + P.
25 25
Answer
Marks
Guidance
Attempt at resolving parallel to the plane to get an equation
*M1
Correct number of relevant terms, allow sign errors, allow sin/cos
mix, allow g missing.
For reference F+Pcos16.26...=1.2gsin16.26...
allow with an angle of 16 or better.
24 7
F+P =1.2g
Answer
Marks
Guidance
25 25
A1
24 84
F+0.96P=3.36 or F+ P= .
25 25
Answer
Marks
Guidance
Use of F=0.15Rto get an equation in P only
DM1
Dependent on both previous M marks – where R is initially a
linear combination of a P component and a weight component (or
a mass component).
7 24 24 7
1.2g −P =0.151.2g +P .
25 25 25 25
Answer
Marks
Guidance
Solve to get P=1.63
A1
272
Allow ,1.62874...
167
6
Answer
Marks
Guidance
Question
Answer
Marks
Question 6:
6 | Attempt at resolving perpendicular to the plane to get an equation | *M1 | Correct number of relevant terms, allow sign errors, allow sin/cos
mix, allow g missing.
For reference R=1.2gcos16.26...+Psin16.26... - allow with
an angle of 16 or better.
24 7
R=1.2g +P
25 25 | A1 | 288 7
R=11.52+0.28P or R= + P.
25 25
Attempt at resolving parallel to the plane to get an equation | *M1 | Correct number of relevant terms, allow sign errors, allow sin/cos
mix, allow g missing.
For reference F+Pcos16.26...=1.2gsin16.26...
allow with an angle of 16 or better.
24 7
F+P =1.2g
25 25 | A1 | 24 84
F+0.96P=3.36 or F+ P= .
25 25
Use of F=0.15Rto get an equation in P only | DM1 | Dependent on both previous M marks – where R is initially a
linear combination of a P component and a weight component (or
a mass component).
7 24 24 7
1.2g −P =0.151.2g +P .
25 25 25 25
Solve to get P=1.63 | A1 | 272
Allow ,1.62874...
167
6
Question | Answer | Marks | Guidance
\includegraphics{figure_6}
A particle of mass $1.2$ kg is placed on a rough plane which is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. The particle is kept in equilibrium by a horizontal force of magnitude $P$ N acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and the plane is $0.15$.
Find the least possible value of $P$. [6]
\hfill \mbox{\textit{CAIE M1 2024 Q6 [6]}}