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A particle of weight 24 N is held in equilibrium by two light inextensible strings - Edexcel - A-Level Maths Mechanics - Question 1 - 2007 - Paper 1

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A particle of weight 24 N is held in equilibrium by two light inextensible strings. One string is horizontal. The other string is inclined at an angle of 30° to the ... show full transcript

Worked Solution & Example Answer:A particle of weight 24 N is held in equilibrium by two light inextensible strings - Edexcel - A-Level Maths Mechanics - Question 1 - 2007 - Paper 1

Step 1

a) the value of P

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Answer

To find the value of P, we first examine the vertical forces acting on the particle. The weight of the particle is given as 24 N. The vertical component of the tension P can be expressed as:

Pimesextsin(30°)P imes ext{sin}(30°)

In equilibrium, the total vertical forces must balance the weight of the particle. Thus, we can set up the equation:

Pimesextsin(30°)=24P imes ext{sin}(30°) = 24

Now substituting the known value of sin(30°), which is 0.5:

Pimes0.5=24P imes 0.5 = 24

Solving for P gives:

P=240.5=48.P = \frac{24}{0.5} = 48.

Step 2

b) the value of Q

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Answer

Next, we calculate the value of Q, which is the horizontal component of the tension in the string at an angle of 30°. The horizontal component can be expressed as:

Q=Pimesextcos(30°)Q = P imes ext{cos}(30°)

Using the value of P we found earlier:

Q=48imesextcos(30°)Q = 48 imes ext{cos}(30°)

Knowing that ( ext{cos}(30°) = \frac{\sqrt{3}}{2} \ ext{ or } 0.866), we can substitute this into our equation:

Q=48imes3241.6.Q = 48 imes \frac{\sqrt{3}}{2} \approx 41.6.

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