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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

(a) the tension in BC

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Answer

To solve for the tension in BC, we start by applying the equilibrium conditions in the horizontal direction. Given that the tension in AC is 20 N:

Using the cosine components:

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

Calculating the values:

cos⁡(30∘)=32,cos⁡(60∘)=12\cos(30^\circ) = \frac{\sqrt{3}}{2}, \cos(60^\circ) = \frac{1}{2}

This leads to:

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

Simplifying gives:

T=203≈34.64 NT = 20 \sqrt{3} \approx 34.64 \, \text{N}

Step 2

(b) the value of m

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Answer

For the vertical equilibrium, we apply the weight balance:

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

Substituting the known values:

g≈10 m/s2,sin⁡(30∘)=12,sin⁡(60∘)=32g \approx 10 \, \text{m/s}^2, \sin(30^\circ) = \frac{1}{2}, \sin(60^\circ) = \frac{\sqrt{3}}{2}

Thus,

m \cdot 10 = 20 \cdot \frac{1}{2} + T \cdot \frac{\sqrt{3}}{2}$$ Substituting for T:

m \cdot 10 = 10 + 34.64 \cdot \frac{\sqrt{3}}{2}

Calculatingfurther:Calculating further:

m \cdot 10 = 10 + 30 \approx 40.64

Finally,Finally,

m = \frac{40.64}{10} \approx 4.064 , \text{kg}$$.

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