A uniform rod \(A B\) rests in limiting equilibrium in a vertical plane with the end \(A\) on rough horizontal ground and the end \(B\) against a rough vertical wall that is perpendicular to the plane of the rod. The angle between the rod and the ground is \(\theta\). The coefficient of friction between the rod and the wall is \(\mu\), and the coefficient of friction between the rod and the ground is \(2 \mu\). Show that \(\tan \theta = \frac { 1 - 2 \mu ^ { 2 } } { 4 \mu }\).
Given that \(\theta \leqslant 45 ^ { \circ }\), find the set of possible values of \(\mu\).
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Question 11:
Part (a):
Answer Marks
Resolve horizontally and vertically: \(F_A = R_B\) B1
Relate limiting frictions and reactions at \(A\) and \(B\): \(F_A = 2\mu R_A\) *and* \(F_B = \mu R_B\) B1
EITHER:
Answer Marks
Resolve vertically: \(F_B + R_A = W\) B1
Combine equations to find \(R_A\): \(R_A = W/(1 + 2\mu^2)\) M1 A1
Take moments about \(B\): \(2R_A \cos\theta - 2F_A \sin\theta = W\cos\theta\) M1 A1
Rearrange and use \(F_A = 2\mu R_A\): \(R_A - 2\mu R_A \tan\theta = \frac{1}{2}W\)
Answer Marks
\(R_A = \frac{1}{2}W/(1 - 2\mu\tan\theta)\) A1
Combine to find \(\tan\theta\): \(\tan\theta = (1 - 2\mu^2)/4\mu\) A.G. M1 A1
OR:
Answer Marks
Resolve vertically: \(F_B + R_A = W\) (B1)
Combine equations to find \(R_B\): \(R_B = 2\mu W/(1 + 2\mu^2)\) (M1 A1)
Take moments about \(A\): \(2R_B \sin\theta + 2F_B \cos\theta = W\cos\theta\) (M1 A1)
Rearrange and use \(F_B = \mu R_B\): \(R_B \tan\theta + \mu R_B = \frac{1}{2}W\)
Answer Marks
\(R_B = \frac{1}{2}W/(\tan\theta + \mu)\) (A1)
Combine to find \(\tan\theta\): \(\tan\theta = (1 - 2\mu^2)/4\mu\) A.G. (M1 A1)
OR:
Answer Marks
Guidance
Take moments about centre of rod: \((F_A + R_B)\sin\theta = (R_A - F_B)\cos\theta\) (M2 A1 A1)
Use first 3 equations to eliminate 3 forces: \(\tan\theta = (R_B/2\mu - \mu R_B)/(R_B + R_B)\) (M1 A1 A1)
\(= (1 - 2\mu^2)/4\mu\) A.G. (A1)
10 marks
Find positive value of \(\mu\) when \(\theta = 45°\): \(2\mu^2 + 4\mu - 1 = 0\)
Answer Marks
Guidance
\(\mu = \sqrt{3/2} - 1\) *or* \(0.225\) A.E.F M1 A1
Find positive value of \(\mu\) when \(\theta = 0°\): \(\mu = 1/\sqrt{2}\) *or* \(0.707\) A.E.F B1
State set of possible values of \(\mu\) (A.E.F.): \([\sqrt{3/2}-1, 1/\sqrt{2}]\) *or* \([0.225, 0.707]\) A1
4 marks
Question 11(b):
Tabulate observed values:
Answer Marks
Guidance
Rural Urban
Support \(A\)
\(60-A\)
Not support \(45-A\)
\(A-5\)
M1 A1
Find corresponding expected values:
Answer Marks
\(27, 33, 18, 22\) M1 A1
Calculate value of \(\chi^2\):
Answer Marks
\(\chi^2 = \frac{(27-A)^2}{27} + \frac{(A-27)^2}{18} + \frac{(A-27)^2}{33} + \frac{(27-A)^2}{22}\) M1
\(= \frac{(50/297)(A-27)^2}{}\)
Answer Marks
*or* \(0.168[35](A-27)^2\) A1
State or use correct tabular \(\chi^2\) value: \(\chi^2_{\text{tab}} = \chi^2_{1,0.9} = 2.706\) (to 3 s.f.) B1
Use conclusion of independence to find equation for \(A\): \(\frac{(50/297)(A-27)^2 < \chi^2_{\text{tab}}}{}\) M1
Find bounds for \(A\) (to 3 s.f.) (integer valued required for A1): \((A-27)^2 < 16.07\)
Answer Marks
Guidance
\(A_{\min} = 23\) *and* \(A_{\max} = 31\) A1, A1
10 marks
Relate new value \(\chi^2_{\text{new}}\) to original \(\chi^2\):
Answer Marks
\(\chi^2_{\text{new}} = N \times \chi^2\) M1
Use conclusion of independence to find equation for \(A\): \(0.168[35]\, N(A-27)^2 < \chi^2_{\text{tab}}\) M1
Find \(N_{\max}\) with \(A = 29\) (integer value required for A1): \(N < 2.706/(4 \times 0.16835) = 4.02\)
Answer Marks
Guidance
\(N_{\max} = 4\) A1, A1
4 marks
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# Question 11:
## Part (a):
Resolve horizontally and vertically: $F_A = R_B$ | B1 |
Relate limiting frictions and reactions at $A$ and $B$: $F_A = 2\mu R_A$ *and* $F_B = \mu R_B$ | B1 |
**EITHER:**
Resolve vertically: $F_B + R_A = W$ | B1 |
Combine equations to find $R_A$: $R_A = W/(1 + 2\mu^2)$ | M1 A1 |
Take moments about $B$: $2R_A \cos\theta - 2F_A \sin\theta = W\cos\theta$ | M1 A1 |
Rearrange and use $F_A = 2\mu R_A$: $R_A - 2\mu R_A \tan\theta = \frac{1}{2}W$
$R_A = \frac{1}{2}W/(1 - 2\mu\tan\theta)$ | A1 |
Combine to find $\tan\theta$: $\tan\theta = (1 - 2\mu^2)/4\mu$ **A.G.** | M1 A1 |
**OR:**
Resolve vertically: $F_B + R_A = W$ | (B1) |
Combine equations to find $R_B$: $R_B = 2\mu W/(1 + 2\mu^2)$ | (M1 A1) |
Take moments about $A$: $2R_B \sin\theta + 2F_B \cos\theta = W\cos\theta$ | (M1 A1) |
Rearrange and use $F_B = \mu R_B$: $R_B \tan\theta + \mu R_B = \frac{1}{2}W$
$R_B = \frac{1}{2}W/(\tan\theta + \mu)$ | (A1) |
Combine to find $\tan\theta$: $\tan\theta = (1 - 2\mu^2)/4\mu$ **A.G.** | (M1 A1) |
**OR:**
Take moments about centre of rod: $(F_A + R_B)\sin\theta = (R_A - F_B)\cos\theta$ | (M2 A1 A1) |
Use first 3 equations to eliminate 3 forces: $\tan\theta = (R_B/2\mu - \mu R_B)/(R_B + R_B)$ | (M1 A1 A1) |
$= (1 - 2\mu^2)/4\mu$ **A.G.** | (A1) | 10 marks |
Find positive value of $\mu$ when $\theta = 45°$: $2\mu^2 + 4\mu - 1 = 0$
$\mu = \sqrt{3/2} - 1$ *or* $0.225$ | A.E.F M1 A1 |
Find positive value of $\mu$ when $\theta = 0°$: $\mu = 1/\sqrt{2}$ *or* $0.707$ | A.E.F B1 |
State set of possible values of $\mu$ (A.E.F.): $[\sqrt{3/2}-1, 1/\sqrt{2}]$ *or* $[0.225, 0.707]$ | A1 | 4 marks | **Total: 14**
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# Question 11(b):
## Tabulate observed values:
| | Rural | Urban |
|---|---|---|
| Support | $A$ | $60-A$ |
| Not support | $45-A$ | $A-5$ |
| M1 A1 |
## Find corresponding expected values:
$27, 33, 18, 22$ | M1 A1 |
## Calculate value of $\chi^2$:
$\chi^2 = \frac{(27-A)^2}{27} + \frac{(A-27)^2}{18} + \frac{(A-27)^2}{33} + \frac{(27-A)^2}{22}$ | M1 |
$= \frac{(50/297)(A-27)^2}{}$
*or* $0.168[35](A-27)^2$ | A1 |
State or use correct tabular $\chi^2$ value: $\chi^2_{\text{tab}} = \chi^2_{1,0.9} = 2.706$ (to 3 s.f.) | B1 |
Use conclusion of independence to find equation for $A$: $\frac{(50/297)(A-27)^2 < \chi^2_{\text{tab}}}{}$ | M1 |
Find bounds for $A$ (to 3 s.f.) (integer valued required for A1): $(A-27)^2 < 16.07$
$A_{\min} = 23$ *and* $A_{\max} = 31$ | A1, A1 | 10 marks |
## Relate new value $\chi^2_{\text{new}}$ to original $\chi^2$:
$\chi^2_{\text{new}} = N \times \chi^2$ | M1 |
Use conclusion of independence to find equation for $A$: $0.168[35]\, N(A-27)^2 < \chi^2_{\text{tab}}$ | M1 |
Find $N_{\max}$ with $A = 29$ (integer value required for A1): $N < 2.706/(4 \times 0.16835) = 4.02$
$N_{\max} = 4$ | A1, A1 | 4 marks | **Total: 14**
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A uniform rod $A B$ rests in limiting equilibrium in a vertical plane with the end $A$ on rough horizontal ground and the end $B$ against a rough vertical wall that is perpendicular to the plane of the rod. The angle between the rod and the ground is $\theta$. The coefficient of friction between the rod and the wall is $\mu$, and the coefficient of friction between the rod and the ground is $2 \mu$. Show that $\tan \theta = \frac { 1 - 2 \mu ^ { 2 } } { 4 \mu }$.
Given that $\theta \leqslant 45 ^ { \circ }$, find the set of possible values of $\mu$.
\hfill \mbox{\textit{CAIE FP2 2013 Q11 EITHER}}