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A particle of mass m kg is attached at C to two light inextensible strings AC and BC - Edexcel - A-Level Maths Mechanics - Question 3 - 2010 - Paper 1

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A particle of mass m kg is attached at C to two light inextensible strings AC and BC. The other ends of the strings are attached to fixed points A and B on a horizon... show full transcript

Worked Solution & Example Answer:A particle of mass m kg is attached at C to two light inextensible strings AC and BC - Edexcel - A-Level Maths Mechanics - Question 3 - 2010 - Paper 1

Step 1

Find the tension in BC

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Answer

To solve for the tension in BC, we start from the equilibrium conditions. The horizontal components of the tensions must balance each other. Hence, we write:

20cos⁡(30∘)=Tcos⁡(60∘)20 \cos(30^\circ) = T \cos(60^\circ)

Substituting the values of cosine:

20⋅32=T⋅1220 \cdot \frac{\sqrt{3}}{2} = T \cdot \frac{1}{2}

Simplifying this leads to:

T=2032⋅12=203≈34.64 NT = \frac{20 \sqrt{3}}{2 \cdot \frac{1}{2}} = 20 \sqrt{3} \approx 34.64 \, \text{N}

Step 2

Find the value of m

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Answer

For the vertical components, the weight of the particle must equal the sum of the vertical components of the tensions:

mg=20sin⁡(30∘)+Tsin⁡(60∘)mg = 20 \sin(30^\circ) + T \sin(60^\circ)

Substituting the tension value we found earlier:

mg=20⋅12+203⋅32mg = 20 \cdot \frac{1}{2} + 20 \sqrt{3} \cdot \frac{\sqrt{3}}{2}

Simplifying gives:

mg=10+30≈40 Nmg = 10 + 30 \approx 40 \, \text{N}

Now, to find m:

m=40gm = \frac{40}{g}

Assuming standard gravity, we can calculate:

m≈409.81≈4.08 kgm \approx \frac{40}{9.81} \approx 4.08 \, \text{kg}

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