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A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall - Edexcel - A-Level Maths Mechanics - Question 9 - 2018 - Paper 1
Question 9
A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The plank is held in a horizontal position by a rope. One end of the rope i... show full transcript
Worked Solution & Example Answer:A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall - Edexcel - A-Level Maths Mechanics - Question 9 - 2018 - Paper 1
Step 1
Using the model, show that the tension in the rope is T = \( \frac{5Mg(3x + α)}{6a} \).
Answer
To find the tension in the rope, we will take moments about point A. The sum of moments around A must equal zero for the system to be in equilibrium:
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The moment due to the weight of the block (3M) about A:
- Moment = Weight × Distance = ( 3Mg \cdot x )
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The moment due to the tension in the rope:
- The vertical component of the tension is ( T \sin(α) )
- Distance from A to the line of action of T = 2a
- Moment = ( T \sin(α) imes 2a )
Setting the sum of moments equal to zero:
[ 3Mgx = T \sin(α) imes 2a ]
Now substituting for ( an(α) ), where ( \tan(α) = \frac{3}{4} ), we need to determine ( rac{ ext{sin}(α)}{ ext{cos}(α)} ):
[ ext{sin}(α) = \frac{3}{5}, \quad ext{cos}(α) = \frac{4}{5} ]
Thus, we have:
[ T \cdot \frac{3}{5} imes 2a = 3Mgx ]
From which, rearranging gives:
[ T = \frac{5Mg(3x + α)}{6a} ]
Step 2
Find x in terms of a.
Answer
From the previous result:
[ T = \frac{5Mg(3x + α)}{6a} ]
We set ( T = 2Mg ) (the maximum tension). Thus:
[ 2Mg = \frac{5Mg(3x + α)}{6a} ]
Cross-multiplying gives:
[ 12a = 5(3x + α) ]
Solving for x results in:
[ x = \frac{12a}{15} - \frac{α}{15} = \frac{2a}{3} - \frac{α}{15} ]
Step 3
The force exerted on the plank at A by the wall acts in a direction which makes an angle β with the horizontal.
Answer
For the force at A exerted by the wall, we apply the equilibrium conditions in the horizontal direction:
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The horizontal component of the force at A will balance the horizontal component of the tension in the rope:
- From previous steps, the tension component in the horizontal direction is given by ( T \cos(α) ).
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Thus:
- The horizontal component at A must counteract this force to maintain equilibrium:
- Force exerted at A = ( T \cos(α) )
Using trigonometric identities:
[ tan(β) = \frac{Y}{X} \text{ (where } Y ext{ is vertical and } X ext{ is horizontal)} ]
Substituting for Y using the maximum tension gives:
[ tan(β) = \frac{5Mg}{2Mg} = \frac{5}{4} ]
Step 4
Explain how this will restrict the possible positions of P. You must justify your answer carefully.
Answer
For the system to maintain equilibrium, the tension T must not exceed 5Mg. From our earlier equations, we find:
[ 5Mg(3x + α) \leq 6a ]
Rearranging gives:
[ x \leq \frac{5a}{3} \text{ when considering other constraints from } A. ]
This means that the block cannot shift too far along the plank; otherwise, the tension in the rope will surpass the limit and could result in the rod breaking. Thus, the position of point P is constrained to ensure that the tension remains within safe limits.