Standard +0.3 This is a standard M1 friction problem requiring resolution of forces in two perpendicular directions (parallel and perpendicular to the plane) and application of F=μR at limiting equilibrium. While it involves multiple force components including an angled string, the method is routine and well-practiced. The calculation is straightforward once the setup is complete, making it slightly easier than average.
\includegraphics{figure_5}
A block of mass \(12\text{kg}\) is placed on a plane which is inclined at an angle of \(24°\) to the horizontal. A light string, making an angle of \(36°\) above a line of greatest slope, is attached to the block. The tension in the string is \(65\text{N}\) (see diagram). The coefficient of friction between the block and plane is \(\mu\). The block is in limiting equilibrium and is on the point of sliding up the plane.
Find \(\mu\). [6]
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.
Answer
Marks
Guidance
5
Attempt at resolving parallel to the plane
*M1
otherwise dimensionally correct.
Answer
Marks
Guidance
65cos3612gsin24F
A1
F 3.777707
Attempt at resolving perpendicular to the plane
*M1
3 terms. Allow sign errors, sin/cos mix. Allow g missing,
otherwise dimensionally correct.
Answer
Marks
Guidance
12gcos24 R65sin36
A1
R71.419
Use F R
65cos3612gsin24 52.58648.808 3.777
Answer
Marks
Guidance
12gcos2465sin36 109.62538.206 71.419
DM1
To get an equation in only.
Dependent on two previous M marks. Allow g missing
Answer
Marks
Guidance
0.0529
A1
Allow AWRT 0.053
Do not accept fractional equivalent.
6
Answer
Marks
Guidance
Question
Answer
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 | Attempt at resolving parallel to the plane | *M1 | 3 terms. Allow sign errors, sin/cos mix. Allow g missing,
otherwise dimensionally correct.
65cos3612gsin24F | A1 | F 3.777707
Attempt at resolving perpendicular to the plane | *M1 | 3 terms. Allow sign errors, sin/cos mix. Allow g missing,
otherwise dimensionally correct.
12gcos24 R65sin36 | A1 | R71.419
Use F R
65cos3612gsin24 52.58648.808 3.777
12gcos2465sin36 109.62538.206 71.419 | DM1 | To get an equation in only.
Dependent on two previous M marks. Allow g missing
0.0529 | A1 | Allow AWRT 0.053
Do not accept fractional equivalent.
6
Question | Answer | Marks | Guidance
\includegraphics{figure_5}
A block of mass $12\text{kg}$ is placed on a plane which is inclined at an angle of $24°$ to the horizontal. A light string, making an angle of $36°$ above a line of greatest slope, is attached to the block. The tension in the string is $65\text{N}$ (see diagram). The coefficient of friction between the block and plane is $\mu$. The block is in limiting equilibrium and is on the point of sliding up the plane.
Find $\mu$. [6]
\hfill \mbox{\textit{CAIE M1 2022 Q5 [6]}}