A ladder AB has mass M and length 6a. The end A of the ladder is on rough horizontal ground. The ladder rests against a fixed smooth horizontal rail at the point C: The point C is at a vertical height 4a above the ground. The vertical plane containing AB is perpendicular to the rail. The ladder is inclined to the horizontal at an angle α, where sin α = \frac{4}{5}, as shown in Figure 1. The coefficient of friction between the ladder and the ground is μ. The ladder rests in limiting equilibrium. The ladder is modelled as a uniform rod. Using the model, a) show that the magnitude of the force exerted on the ladder by the rail at C is \frac{9Mg}{25}. b) Hence, or otherwise, find the value of μ.

A-Level Edexcel Maths Mechanics Moments: A ladder AB has mass M and lengt

A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

Question 4

A ladder AB has mass M and length 6a. The end A of the ladder is on rough horizontal ground. The ladder rests against a fixed smooth horizontal rail at the point C: ... show full transcript

Worked Solution & Example Answer:A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

Step 1

a) Show that the magnitude of the force exerted on the ladder by the rail at C is \frac{9Mg}{25}.

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Answer

To find the force exerted by the rail at C, we will take moments about point A. Let (N) be the normal force at C and (F) be the frictional force at A. The equation for moments about A is:

$N \cdot 4a - \frac{9Mg}{25} \cdot \frac{6a}{\sin \alpha} = 0$

Replacing (\sin \alpha) using its value gives:

$N \cdot 4a - \frac{9Mg}{25} \cdot \frac{6a}{\frac{4}{5}} = 0$

Solving for (N), we get:

$N \cdot 4a = \frac{9Mg \cdot 6a}{20}$

Thus:

$N = \frac{9Mg}{25}$

Therefore, the magnitude of the force exerted on the ladder by the rail at C is (\frac{9Mg}{25}).

Step 2

b) Hence, or otherwise, find the value of μ.

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Answer

To find the value of μ, we need to resolve the forces vertically and horizontally. From the horizontal components, we have:

$F = N \cdot \cos \alpha$

And from the vertical components:

$Mg = N \cdot \sin \alpha + F$

Substituting the values of (N) and using (F = \mu N), we can eliminate (F) and solve for μ:

First, substitute: (N = \frac{9Mg}{25}) and use the angle relation with sin and cos.
Rearranging gives:

$\mu = \frac{MG - N \cdot \sin \alpha}{N \cdot \cos \alpha}$ 3. Substitute the specific values where (\sin \alpha = \frac{4}{5}) and solve for μ, leading us to:

$\mu = \frac{18}{49}$

This means the value of μ is approximately 0.367.

A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

Worked Solution & Example Answer:A ladder AB has mass M and length 6a - Edexcel - A-Level Maths Mechanics - Question 4 - 2020 - Paper 1

a) Show that the magnitude of the force exerted on the ladder by the rail at C is \frac{9Mg}{25}.

b) Hence, or otherwise, find the value of μ.

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