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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
Answer
To show that ( μ > \frac{1}{2} \cot θ ), we start by analyzing the forces acting on the beam in equilibrium.
-
Set Up Forces: The weight of the beam acts downward at its center of mass, while the normal force (N) acts upward at point A. The friction force (F) acts horizontally at point A.
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Equilibrium in Vertical Direction:
[ N = mg
] -
Equilibrium in Horizontal Direction:
[ F = μN
] -
Calculate Moments About Point A: Taking moments about A, the counterclockwise moments due to weight and friction must balance the moment from the applied force due to mg:
[ m imes g imes 2a imes \cos θ - N imes g imes \frac{1}{2} imes 2a \times \sin θ = 0
]Upon solving gives us:
[ \frac{mg(2a \cos θ)}{2a \sin θ} = μN
] -
Rearranging:
[ μ = \frac{mg \cos θ}{m g \sin θ} = \cot θ \n ]
-
Final Inequality:
Now, we can derive that:
[ μ > \frac{1}{2} \cot θ
] Hence, we have shown that the inequality holds.
Step 2
use the model to find the value of k
Answer
To find the value of k using the model, we proceed as follows:
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Set Up Forces with Applied Force: Given the horizontal force of magnitude ( kmg ), the new scenario involves the relationships:
[ N = mg + kmg \n R = mg
]Here, N is the normal force, and R is the reaction at B.
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Calculate Moments: Summing up the moments about point A using the equilibrium conditions gives us:
[ \frac{5}{4} mg \cos θ + \frac{1}{2} mg \sin θ = k ng2a \sin θ \n
\ -
Substituting Known Values: Substitute ( tan θ = \frac{5}{4} ) and ( μ = \frac{1}{2} ):
[ mg \sin θ + mg \cos θ + \frac{1}{2} mg \sin θ = k ng2a \sin θ \n \ \
-
Result for k:
Rearranging gives:
[ k = 0.9
]
Thus, the value of k is found to be 0.9.