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A bench consists of a plank which is resting in a horizontal position on two thin vertical legs - Edexcel - A-Level Maths Mechanics - Question 4 - 2009 - Paper 1
Question 4
A bench consists of a plank which is resting in a horizontal position on two thin vertical legs. The plank is modelled as a uniform rod PS of length 2.4 m and mass 2... show full transcript
Worked Solution & Example Answer:A bench consists of a plank which is resting in a horizontal position on two thin vertical legs - Edexcel - A-Level Maths Mechanics - Question 4 - 2009 - Paper 1
Step 1
(a) the magnitude of the normal reaction between the plank and the leg at Q and the magnitude of the normal reaction between the plank and the leg at R.
Answer
To find the normal reactions at Q and R, we consider the equilibrium of the plank as it is in rest:
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Calculate the total weight supported:
The total weight of the plank plus the weights of Arthur and Beatrice is computed as follows:
Total mass = Mass of plank + Mass of Arthur + Mass of Beatrice = 20 ext{ kg} + 60 ext{ kg} + 40 ext{ kg} = 120 ext{ kg}
Therefore, the total weight (W) is:
W = 120 ext{ kg} imes 9.81 ext{ m/s}^2 = 1177.2 ext{ N}.
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Considering moments about point Q:
Let the normal reaction at Q be C and at R be D. By taking moments about Q,
60 g imes 0.8 + 40 g imes 2.0 - D imes 1.6 = 0.
Where
- 60 g = weight of Arthur = 60 * 9.81 N = 588.6 N
- 40 g = weight of Beatrice = 40 * 9.81 N = 392.4 N
Solving leads to:
D = 30 g.
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Applying the vertical forces equilibrium:
C + D = 120 g,
Thus, substituting D:
C + 30 g = 120 g ightarrow C = 90 g.
Thus, the magnitudes of the normal reactions are:
- Normal reaction at Q (C) is 90 g.
- Normal reaction at R (D) is 30 g.
Step 2
(b) find the distance QX.
Answer
As Beatrice remains at point P and Arthur moves to point X, we know:
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Establishing the new conditions:
Since the normal reaction at Q is now twice that of R:
Let the reaction at R be F, then the reaction at Q is 2F.
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Setting up the moment equation about Q:
2F + F = 40g + 60g + 20g,
Where the weights included are from Beatrice, Arthur, and the plank.
Therefore, solving,
3F = 120g ightarrow F = 40g.
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Finding Reaction at Q:
Reaction at Q is 2F = 80g.
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Apply the moments about point Q again:
The weights contribute to the moments:
60g imes (0.8 + x) + 20g imes 0.4 = 40g imes 0.4 + 80g imes 0.4.
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Solving for x:
Rearranging gives:
Rearranging leads to:
x = rac{16}{15} = 1.07 m.
Thus, the distance QX is 1.07 m.