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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 start by analyzing the forces acting on the particle in vertical equilibrium. The weight of the particle is given as 24 N. Since the tension in the string inclined at 30° is P, we can use the vertical component of this tension to balance the weight.

The vertical component can be represented as: Pimesextsin(30°)=24P imes ext{sin}(30°) = 24

Solving for P: P=24sin(30°)P = \frac{24}{\text{sin}(30°)}

Since (\text{sin}(30°) = \frac{1}{2}), we can substitute: P=2412=24×2=48P = \frac{24}{\frac{1}{2}} = 24 \times 2 = 48

Thus, the value of P is 48 N.

Step 2

b) the value of Q

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Answer

Next, to find the value of Q, we will use the horizontal equilibrium condition. The only horizontal force acting on the particle is due to the tension in the horizontal string, which is Q, and the horizontal component of the tension P.

Using trigonometric relationships: Q=P×cos(30°)Q = P \times \text{cos}(30°)

Substituting the known value of P: Q=48×cos(30°)Q = 48 \times \text{cos}(30°)

Given that (\text{cos}(30°) = \frac{\sqrt{3}}{2}): Q=48×32=243Q = 48 \times \frac{\sqrt{3}}{2} = 24\sqrt{3}

Numerically, this approximates to: Q41.6Q \approx 41.6

Thus, the value of Q is approximately 41.6 N.

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