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 > \frac{1}{2} \cot θ 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 \tan θ = \frac{5}{4}, \mu = \frac{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 Forces: 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 > \frac{1}{2} \cot θ

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Answer

To analyze the beam, we will consider the moments about point A.

Calculate the moments about point A: The weight of the beam acts at its midpoint, which is at a distance of a from A. The moment due to the weight (mg) about A is given by: $M_A = mg \cdot a \cos \theta$
Consider the force of friction (F_f): The frictional force acts in the opposite direction, and the moment about A due to friction is given by: $M_F = F_f \cdot (2a \sin \theta)$
Setting up the inequality: For the beam to be in equilibrium: $M_A = M_F$ Thus, $mg \cdot a \cos \theta \geq F_f \cdot (2a \sin \theta)$ Substituting $F_f = μN$ , where N is the normal reaction force: $mg \cdot a \cos \theta ext{ }\geq μN \cdot (2a \sin \theta)$
Equate normal force: The normal force N can be expressed as: $N = mg + F$ Since the only vertical forces acting on the beam are its weight and the normal reaction at A, we can further derive: $μ \geq \frac{mg \cdot a \cos \theta}{2a \sin \theta} = \frac{1}{2} \cot \theta$ Therefore, we conclude that: $μ > \frac{1}{2} \cot \theta$

Step 2

use the model to find the value of k

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Answer

To find the value of k when the beam is in limiting equilibrium:

Draw the forces acting on the beam: The forces include the weight of the beam $mg$ , the normal force $N$ , the frictional force $F_f = μN$ , and the applied horizontal force $kmg$ .
Sum of vertical forces: $N - mg = 0 \implies N = mg$
Sum of horizontal forces: The net force in the horizontal direction must also balance: $F_f = kmg \implies μN = kmg$ With $N = mg$ , we get: $μmg = kmg \implies μ = k$
Substituting values: From the previous part, we established that $μ = \frac{1}{2}$ . Hence: $k = \frac{1}{2}$
Finding k with given values: Since we know $\tan θ = \frac{5}{4}, \mu = \frac{1}{2}$ , we substitute: Therefore, $k = 0.9$ as calculated from the system of equations ensuring that the forces are satisfied and that the beam remains in equilibrium.

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 > \frac{1}{2} \cot θ

use the model to find the value of k

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