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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

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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... 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

Part (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

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Answer

To find the magnitudes of the normal reactions at points QQ and RR, we begin by applying the equilibrium conditions for forces and moments on the plank.

Let:

  • CC = normal reaction at QQ
  • DD = normal reaction at RR

Step 1: Sum of forces The total downward force due to weights is given by: C+D=120extNC + D = 120 ext{N} (where weight of the plank plus weights of Arthur and Beatrice equals 120N)

Step 2: Moments about one end of the plank Taking moments about point QQ: M(Q)=80imes0.8+40imes0.440imes0.420imes0.4=0M(Q) = 80 imes 0.8 + 40 imes 0.4 - 40 imes 0.4 - 20 imes 0.4 = 0
This simplifies to (Dimes1.6=80imes0.8+40imes0.4D imes 1.6 = 80 imes 0.8 + 40 imes 0.4), which can be solved to yield: D=30extND = 30 ext{N}

Step 3: Solving for C: Substituting DD back into the first equation gives:

ightarrow C = 90 ext{N}$$ The magnitudes of the normal reactions are: - $C = 90 ext{N}$ - $D = 30 ext{N}$

Step 2

Part (b): find the distance QX

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Answer

After Beatrice remains at PP, Arthur will now sit at point XX, and we know that the normal reaction at QQ is now twice that of RR.

Let F=2DF = 2D (the normal reaction at QQ).

Step 1: New upward force equation 2F+F=40+20+602F + F = 40 + 20 + 60
Thus: 3F=120extN3F = 120 ext{N} This gives: F=40extNF = 40 ext{N} and therefore: D=20extND = 20 ext{N}.

Step 2: New moments about point Q: M(Q)=60imes0.4+20imes0.4+40imes(2.4x)=40imes0.4M(Q) = 60 imes 0.4 + 20 imes 0.4 + 40 imes (2.4-x) = 40 imes 0.4
Solving this will give: 60x+20(0.4)+40(2.4x)=060x + 20(0.4) + 40(2.4 - x) = 0

On simplifying, we find: x = rac{16}{15} ext{ m} ext{ (or approximately 1.07 m)}

Therefore, the distance QXQX is approximately 1.07 m.

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