Pre-U Pre-U 9795/2 2016 June — Question 10 8 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2016
SessionJune
Marks8
TopicMoments
TypeLadder on smooth wall and rough ground
DifficultyStandard +0.8 This is a two-part ladder equilibrium problem requiring moments, friction, and resolution of forces. Part (i) involves finding limiting equilibrium angle using moment equation about a strategic point, while part (ii) requires finding maximum distance before slipping occurs. The problem demands careful setup of multiple equations (moments, vertical/horizontal equilibrium, friction law) and algebraic manipulation, which is more challenging than routine statics questions but follows standard ladder problem methodology taught in mechanics courses.
Spec3.04b Equilibrium: zero resultant moment and force

10 A uniform ladder \(A B\) of length 5 m and mass 8 kg is placed at an angle \(\theta\) to the horizontal, with \(A\) on rough horizontal ground and \(B\) against a smooth vertical wall. The coefficient of friction between the ladder and the ground is 0.4 .
  1. By taking moments, find the smallest value of \(\theta\) for which the ladder is in equilibrium.
  2. A man of mass 75 kg stands on the ladder when \(\theta = 60 ^ { \circ }\). Find the greatest distance from \(A\) that he can stand without the ladder slipping.

(i) \(N = 80\)
\(M(X)\): \(F\times5\sin\theta = 80\times2.5\cos\theta\)
\(F \leq 0.4N\)
\(\tan\theta \geq 1.25\)
\(\theta_{\min} = \mathbf{51.3°}\) or \(51.4°\), \(0.896\)
- B1: Normal force at ground (can be implied)
- M1: Moments about any point, needs both cos and sin
- M1: Use \(F \leq \mu N\) or \(F = \mu N\)
- M1: Solve equations to obtain \(\tan\theta\)
- A1: Correct answer, in range \([51.3, 51.4]\) or a.r.t. \(0.896\)
[5]
(ii) \(F\times5\sin\theta = (80\times2.5 + 750d)\cos\theta\)
Use \(60°\) and \(\mu\) to obtain
\(d_{\max} = \mathbf{3.57}\)
- M1*: Moments equation with variable \(d\) \([F = 332]\); \([332\times5\times0.5 = 100\sqrt{3} + 337\sqrt{3}\,d]\)
- depM1
- A1: Answer, a.r.t. 3.57 or 3.56
[3]
(i) $N = 80$

$M(X)$: $F\times5\sin\theta = 80\times2.5\cos\theta$

$F \leq 0.4N$

$\tan\theta \geq 1.25$

$\theta_{\min} = \mathbf{51.3°}$ or $51.4°$, $0.896$

- B1: Normal force at ground (can be implied)
- M1: Moments about any point, needs both cos and sin
- M1: Use $F \leq \mu N$ or $F = \mu N$
- M1: Solve equations to obtain $\tan\theta$
- A1: Correct answer, in range $[51.3, 51.4]$ or a.r.t. $0.896$

**[5]**

(ii) $F\times5\sin\theta = (80\times2.5 + 750d)\cos\theta$

Use $60°$ and $\mu$ to obtain

$d_{\max} = \mathbf{3.57}$

- M1*: Moments equation with variable $d$ $[F = 332]$; $[332\times5\times0.5 = 100\sqrt{3} + 337\sqrt{3}\,d]$
- depM1
- A1: Answer, a.r.t. 3.57 or 3.56

**[3]**
10 A uniform ladder $A B$ of length 5 m and mass 8 kg is placed at an angle $\theta$ to the horizontal, with $A$ on rough horizontal ground and $B$ against a smooth vertical wall. The coefficient of friction between the ladder and the ground is 0.4 .\\
(i) By taking moments, find the smallest value of $\theta$ for which the ladder is in equilibrium.\\
(ii) A man of mass 75 kg stands on the ladder when $\theta = 60 ^ { \circ }$. Find the greatest distance from $A$ that he can stand without the ladder slipping.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q10 [8]}}