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A particle P of mass 2 kg is attached to one end of a light string, the other end of which is attached to a fixed point O - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 1

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A particle P of mass 2 kg is attached to one end of a light string, the other end of which is attached to a fixed point O. The particle is held in equilibrium, with ... show full transcript

Worked Solution & Example Answer:A particle P of mass 2 kg is attached to one end of a light string, the other end of which is attached to a fixed point O - Edexcel - A-Level Maths Mechanics - Question 3 - 2013 - Paper 1

Step 1

(i) the value of F

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Answer

To find the value of the force FF, we can use the equilibrium conditions of forces acting on the particle P.

First, we consider the vertical components of the forces. The tension in the string TT can be resolved vertically:

Tcos(30°)+Fcos(60°)=2gT \cos(30°) + F \cos(60°) = 2g

Where g=9.8m/s2g = 9.8 \, \text{m/s}^2, giving us:

  1. Substitute gg: Tcos(30°)+Fcos(60°)=2×9.8T \cos(30°) + F \cos(60°) = 2 \times 9.8 Tcos(30°)+F0.5=19.6T \cos(30°) + F \cdot 0.5 = 19.6

Next, horizontally, the only horizontal component is due to the force FF:

  1. Setting horizontal components: Tcos(30°)Fcos(30°)=0T \cos(30°) - F \cos(30°) = 0 Rearranging gives us: Tcos(30°)=Fcos(30°)T \cos(30°) = F \cos(30°) Thus, we have: F=TF = T

Next, we can substitute this back into the first equation to solve for FF. By combining these equations, we can isolate FF.

Step 2

(ii) the tension in the string

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Answer

To find the tension TT in the string, we will use the values calculated previously.

Substituting the known values: T=2gcos(30°)T = 2g \cos(30°)

Substituting g=9.8g = 9.8: T=2×9.8×cos(30°)T = 2 \times 9.8 \times \cos(30°) Where ( \cos(30°) = \frac{\sqrt{3}}{2} ) yields: T=2×9.8×32=9.8317.0 NT = 2 \times 9.8 \times \frac{\sqrt{3}}{2} = 9.8\sqrt{3} ≈ 17.0\text{ N}

Thus, the tension in the string is approximately 17.0 N.

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