A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{3}$. The vertex B is in contact with a rough horizontal surface for which the coefficient of friction is $\mu$. The lamina is supported by a smooth peg at the point E on AB, where BE = $3a$ (see diagram). (i) Find expressions in terms of $W$ for the normal reaction forces at E and B. (ii) Given that the lamina is about to slip, find the value of $\mu$.

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A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{3}$ - CIE - A-Level Further Maths - Question 2 - 2019 - Paper 1

Question 2

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A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \the... show full transcript

Worked Solution & Example Answer:A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{3}$ - CIE - A-Level Further Maths - Question 2 - 2019 - Paper 1

Step 1

Find expressions in terms of W for the normal reaction forces at E and B.

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Answer

To find the normal reaction forces at points E and B, we can start by taking moments about point B.

The distance BE is given as $3a$ and the weight W acts downwards through the center of gravity of the lamina, which is at a distance of $2a$ from B horizontally.
The moment about point B due to the weight W is: $M_W = W \cdot 2a \cos(\theta)$
The normal reaction force at E, denoted as $R_E$ , will act perpendicular to the surface, thus: $R_E = \frac{W \cdot 2a \cos(\theta)}{3a} = \frac{2W}{3} \cos(\theta)$
Resolving vertically, the sum of the forces must equal zero: $R_B + R_E - W = 0$ Thus, substituting $R_E$ gives: $R_B + \frac{2W}{3} \cos(\theta) = W$ So, $R_B = W - \frac{2W}{3} \cos(\theta) = W\left(1 - \frac{2}{3} \cos(\theta)\right).$

Step 2

Given that the lamina is about to slip, find the value of μ.

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Answer

When the lamina is about to slip, the frictional force at point B can be expressed in terms of the normal reaction and the coefficient of friction:

The frictional force at B is given by: $f = \mu R_B$
This frictional force must be equal to the horizontal component of the forces acting at that point. The forces in the horizontal direction give: $f = R_E \sin(\theta)$
Thus, we can set the expressions equal: $\mu R_B = R_E \sin(\theta)$
Substituting $R_E$ and $R_B$ , we get: $\mu \left(W\left(1 - \frac{2}{3} \cos(\theta)\right)\right) = \frac{2W}{3} \cos(\theta) \sin(\theta).$
Simplifying this will yield the value of $\, \mu$ : $\mu = \frac{\frac{2}{3}\cos(\theta) \sin(\theta)}{W(1 - \frac{2}{3} \cos(\theta))}$ .

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A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{3}$ - CIE - A-Level Further Maths - Question 2 - 2019 - Paper 1

Worked Solution & Example Answer:A uniform square lamina ABCD of side $4a$ and weight $W$ rests in a vertical plane with the edge AB inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac{1}{3}$ - CIE - A-Level Further Maths - Question 2 - 2019 - Paper 1

Find expressions in terms of W for the normal reaction forces at E and B.

Given that the lamina is about to slip, find the value of μ.

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