Pre-U Pre-U 9795/2 2018 June — Question 12 21 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2018
SessionJune
Marks21
TopicMoments
TypeLadder or rod with friction at both contacts
DifficultyStandard +0.8 This is a two-part moments question requiring careful force resolution and equilibrium conditions. Part (i) is straightforward with perpendicular rod setup. Part (ii) is significantly harder: horizontal rod on two inclined planes with limiting friction at both contacts requires simultaneous resolution in two directions, moments equation, and algebraic manipulation to derive the given relationship with an irrational constant. The multi-constraint system and algebraic proof elevate this above typical A-level mechanics.
Spec6.04e Rigid body equilibrium: coplanar forces
12 A uniform \(\operatorname { rod } A B\) has mass 5 kg and length 4 m .
  1. \includegraphics[max width=\textwidth, alt={}, center]{09939c3a-7829-4784-8e6d-ee5356c22cd7-5_529_540_995_840} The rod rests with \(A\) on a rough plane that makes an angle of \(60 ^ { \circ }\) to the horizontal. A string is attached to \(B\) and the rod is in equilibrium in the vertical plane containing the line of greatest slope of the plane, with the string vertical and \(A B\) perpendicular to the plane (see diagram). Find the magnitude of the frictional force at \(A\) and the tension in the string.
  2. \includegraphics[max width=\textwidth, alt={}, center]{09939c3a-7829-4784-8e6d-ee5356c22cd7-5_323_637_1850_794} The rod now rests horizontally with \(A\) in contact with a rough plane that makes an angle of \(60 ^ { \circ }\) with the horizontal and \(B\) in contact with a rough plane that makes an angle of \(30 ^ { \circ }\) with the horizontal (see diagram). The rod and the lines of greatest slope of the two planes are all in the same vertical plane. The coefficients of friction at \(A\) and \(B\) are \(\mu _ { A }\) and \(\mu _ { B }\) respectively. Friction is limiting at both \(A\) and \(B\), with \(A\) on the point of slipping downwards. Show that \(\mu _ { B } = \frac { 1 - \alpha \mu _ { A } } { \alpha + \mu _ { A } }\) where \(\alpha\) is an irrational number to be found.
Question 12(i)
- M(\(B\)): \(4F = 5g\times2\sin60°\)
- M(\(A\)): \(2\times5g\cos30° = 4\times T\cos30°\)
- M(\(C\)): \(2\times F = 2\times T\cos30°\)
- NII(\(\perp AB\)): \(F + T\cos30° = 5g\cos30°\)
- (NII(\(/\!/AB\)): \(T\sin30° + N = 5g\sin30°\))
- (NII(\(\leftrightarrow\)): \(N\cos30° = F\sin30°\))
- (NII(\(\uparrow\)): \(N\sin30° + F\cos30° + T = 5g\)) M3 M1 for attempt to take moments and attempt to derive one other equation by NII or moments about another point; M1 for deriving one useful equation; M1 for deriving a second useful equation (or 2nd and 3rd if \(N\) is involved)
- \(F = 12.5\sqrt{3} = 21.65\) N cwo A1 \([21.6, 21.7]\) or \(12.5\sqrt{3}\) oe
- \(T = 25\) N cwo A1 25 or awrt 25.0
Question 12(ii)
- M1 dep for attempt at moments equation
- One of: M(\(C\)): \(2N_B\cos30° = 2N_A\cos60° + 2F_A\cos30° + 2F_B\cos60°\); M(\(A\)): \(2W + 4F_B\sin30° = 4N_B\sin60°\); M(\(B\)): \(2W = 4N_A\sin30° + 4F_A\sin60°\) A1 one correct moments equation
- Then completing either: Both of M(\(C\)) and NII(\(\leftrightarrow\)): \(N_A\sin60° = F_A\sin30° + N_B\sin30° + F_B\sin60°\); or Three of: One of M(\(C\)) & NII(\(\leftrightarrow\)), M(\(A\)) and/or M(\(B\)), NII(\(//\Pi_A\)): \(F_A + N_B = W\cos30°\), NII(\(\perp\Pi_A\)): \(N_A = F_B + W\sin30°\), NII(\(\uparrow\)): \(W + F_B\cos60° = N_A\cos60° + F_A\cos30° + N_B\cos30°\) M1 Dep on first M1 for other equation(s) to complete
- \(F_A = \mu_A N_A\) and \(F_B = \mu_B N_B\) B1 Can be on diagram or in working
- \(N_B(\sqrt{3}-\mu_B) = N_A(1+\sqrt{3}\mu_A)\); \(N_B(1+\sqrt{3}\mu_B) = N_A(\sqrt{3}-\mu_A)\) M1 Dep on second M1 for simplifying to 2 sim equations in 2 eliminatable 'unknowns' oe elimination
- \(\frac{\sqrt{3}-\mu_B}{1+\sqrt{3}\mu_B} = \frac{1+\sqrt{3}\mu_A}{\sqrt{3}-\mu_A}\) M1 Dep on second M1 for obtaining single equation in \(\mu_A\) and \(\mu_B\) only
- \(\sqrt{3}(\mu_A+\mu_B) + \mu_A\mu_B = 1\) M1 Dep on second M1 for making \(\mu_B\) subject of formula and using exact values of cos/sin
- \(\mu_B = \frac{1-\sqrt{3}\mu_A}{\sqrt{3}+\mu_A}\) A1 final answer in given form, cwo (\(\alpha = \sqrt{3}\)). Note: Form of answer given so must see full working with factorisation leading to single term in \(\mu_B\) for final M1A1. Equation must have \(\mu_A\), \(\mu_B\) and \(\mu_A\mu_B\) terms for M1.
**Question 12(i)**

- M($B$): $4F = 5g\times2\sin60°$
- M($A$): $2\times5g\cos30° = 4\times T\cos30°$
- M($C$): $2\times F = 2\times T\cos30°$
- NII($\perp AB$): $F + T\cos30° = 5g\cos30°$
- (NII($/\!/AB$): $T\sin30° + N = 5g\sin30°$)
- (NII($\leftrightarrow$): $N\cos30° = F\sin30°$)
- (NII($\uparrow$): $N\sin30° + F\cos30° + T = 5g$) **M3** **M1** for attempt to take moments and attempt to derive one other equation by NII or moments about another point; **M1** for deriving one useful equation; **M1** for deriving a second useful equation (or 2nd and 3rd if $N$ is involved)

- $F = 12.5\sqrt{3} = 21.65$ N cwo **A1** $[21.6, 21.7]$ or $12.5\sqrt{3}$ oe

- $T = 25$ N cwo **A1** 25 or awrt 25.0

**Question 12(ii)**

- **M1** dep for attempt at moments equation

- One of: M($C$): $2N_B\cos30° = 2N_A\cos60° + 2F_A\cos30° + 2F_B\cos60°$; M($A$): $2W + 4F_B\sin30° = 4N_B\sin60°$; M($B$): $2W = 4N_A\sin30° + 4F_A\sin60°$ **A1** one correct moments equation

- Then completing either: Both of M($C$) and NII($\leftrightarrow$): $N_A\sin60° = F_A\sin30° + N_B\sin30° + F_B\sin60°$; **or** Three of: One of M($C$) & NII($\leftrightarrow$), M($A$) and/or M($B$), NII($//\Pi_A$): $F_A + N_B = W\cos30°$, NII($\perp\Pi_A$): $N_A = F_B + W\sin30°$, NII($\uparrow$): $W + F_B\cos60° = N_A\cos60° + F_A\cos30° + N_B\cos30°$ **M1** Dep on first M1 for other equation(s) to complete

- $F_A = \mu_A N_A$ and $F_B = \mu_B N_B$ **B1** Can be on diagram or in working

- $N_B(\sqrt{3}-\mu_B) = N_A(1+\sqrt{3}\mu_A)$; $N_B(1+\sqrt{3}\mu_B) = N_A(\sqrt{3}-\mu_A)$ **M1** Dep on second M1 for simplifying to 2 sim equations in 2 eliminatable 'unknowns' oe elimination

- $\frac{\sqrt{3}-\mu_B}{1+\sqrt{3}\mu_B} = \frac{1+\sqrt{3}\mu_A}{\sqrt{3}-\mu_A}$ **M1** Dep on second M1 for obtaining single equation in $\mu_A$ and $\mu_B$ only

- $\sqrt{3}(\mu_A+\mu_B) + \mu_A\mu_B = 1$ **M1** Dep on second M1 for making $\mu_B$ subject of formula and using exact values of cos/sin

- $\mu_B = \frac{1-\sqrt{3}\mu_A}{\sqrt{3}+\mu_A}$ **A1** final answer in given form, cwo ($\alpha = \sqrt{3}$). Note: Form of answer given so must see full working with factorisation leading to single term in $\mu_B$ for final **M1A1**. Equation must have $\mu_A$, $\mu_B$ and $\mu_A\mu_B$ terms for **M1**.
12 A uniform $\operatorname { rod } A B$ has mass 5 kg and length 4 m .\\
(i)\\
\includegraphics[max width=\textwidth, alt={}, center]{09939c3a-7829-4784-8e6d-ee5356c22cd7-5_529_540_995_840}

The rod rests with $A$ on a rough plane that makes an angle of $60 ^ { \circ }$ to the horizontal. A string is attached to $B$ and the rod is in equilibrium in the vertical plane containing the line of greatest slope of the plane, with the string vertical and $A B$ perpendicular to the plane (see diagram). Find the magnitude of the frictional force at $A$ and the tension in the string.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{09939c3a-7829-4784-8e6d-ee5356c22cd7-5_323_637_1850_794}

The rod now rests horizontally with $A$ in contact with a rough plane that makes an angle of $60 ^ { \circ }$ with the horizontal and $B$ in contact with a rough plane that makes an angle of $30 ^ { \circ }$ with the horizontal (see diagram). The rod and the lines of greatest slope of the two planes are all in the same vertical plane. The coefficients of friction at $A$ and $B$ are $\mu _ { A }$ and $\mu _ { B }$ respectively. Friction is limiting at both $A$ and $B$, with $A$ on the point of slipping downwards. Show that $\mu _ { B } = \frac { 1 - \alpha \mu _ { A } } { \alpha + \mu _ { A } }$ where $\alpha$ is an irrational number to be found.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2018 Q12 [21]}}
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