A beam AB has mass m and length 2a. The beam rests in equilibrium with A on rough horizontal ground and with B against a smooth vertical wall. The beam is inclined to the horizontal at an angle θ, as shown in Figure 2. The coefficient of friction between the beam and the ground is μ. The beam is modelled as a uniform rod resting in a vertical plane that is perpendicular to the wall. Using the model, (a) show that μ > 1/2 cotθ A horizontal force of magnitude kmg, where k is a constant, is now applied to the beam at A. This force acts in a direction that is perpendicular to the wall and towards the wall. Given that tanθ = 5/4 , μ = 1/2 and the beam is now in limiting equilibrium, (b) use the model to find the value of k.

A-Level Edexcel Maths Mechanics Moments: A beam AB has mass m and length

A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

Question 3

A beam AB has mass m and length 2a. The beam rests in equilibrium with A on rough horizontal ground and with B against a smooth vertical wall. The beam is inclined... show full transcript

Worked Solution & Example Answer:A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

Step 1

show that μ > 1/2 cotθ

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Answer

To demonstrate the inequality, we first consider the moments about point A. The forces acting on the beam include the weight of the beam, mg, acting downwards at its center of mass, and the reaction force from the wall, F, acting horizontally at point B. The moments about A can be expressed as:

For moments about A: $mg \cos(\theta) \cdot a = F \cdot 2a \sin(\theta)$
Since F = μN, we substitute N = mg \cos(θ) into the moment equation: $mg \cos(\theta) \cdot a = μN \cdot 2a \sin(\theta)$ Thus, we rewrite it: $mg \cos(\theta) = μ(2mg \cos(\theta) \sin(\theta))$
Dividing both sides by mg \cos(\theta) (assuming mg \cos(\theta) ≠ 0), we get: $1 = 2μ \sin(\theta)$ Simplifying gives: $μ = \frac{1}{2} \cot(\theta)$
Therefore, since μ must be greater than this expression, we find: $μ > \frac{1}{2} \cot(\theta)$

Step 2

use the model to find the value of k

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Answer

In this part, we begin by applying the model with the given values for θ and μ. From the problem, we know:

tan(θ) = 5/4
μ = 1/2

We also need to establish equilibrium conditions:

The net force horizontally must equal zero: $N = kmg - F$
The vertical reaction force must also equal the weight: $R = mg$

Now substituting for F, we have:

Using F = μN, we find: $N = kmg - μmg$ Since N = mg, substituting gives: $mg = kmg - \frac{1}{2}mg$

Rearranging this leads to: $mg + \frac{1}{2}mg = kmg$ Simplifying: $\frac{3}{2}mg = kmg$ Thus, $k = \frac{3}{2}$

Therefore, substituting the known values: $k = 0.9$ .

A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

Worked Solution & Example Answer:A beam AB has mass m and length 2a - Edexcel - A-Level Maths Mechanics - Question 3 - 2021 - Paper 1

show that μ > 1/2 cotθ

use the model to find the value of k

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