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Two forces P and Q act on a particle at a point O - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

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Two forces P and Q act on a particle at a point O. The force P has magnitude 15 N and the force Q has magnitude X newtons. The angle between P and Q is 150°, as show... show full transcript

Worked Solution & Example Answer:Two forces P and Q act on a particle at a point O - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Step 1

(a) the magnitude of R

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Answer

To find the magnitude of the resultant force R, we can use the sine rule. According to the problem, we have:

rac{15}{ ext{sin}(50^ ext{o})} = \frac{R}{ ext{sin}(30^ ext{o})}

Rearranging gives:

R=15sin(30exto)sin(50exto)R = 15 \cdot \frac{\text{sin}(30^ ext{o})}{\text{sin}(50^ ext{o})}

Calculating the values:

  • sin(30exto)=0.5\text{sin}(30^ ext{o}) = 0.5
  • sin(50exto)0.7660\text{sin}(50^ ext{o}) \approx 0.7660

Thus,

R150.50.76609.79NR \approx 15 \cdot \frac{0.5}{0.7660} \approx 9.79 \, \text{N}

Step 2

(b) the value of X

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Answer

Next, to find the value of the force Q (magnitude X), we apply the cosine rule:

R2=X2+1522X15cos(100exto)R^2 = X^2 + 15^2 - 2 \cdot X \cdot 15 \cdot \text{cos}(100^ ext{o})

Substituting the known values:

  • From (a), we have R9.79R \approx 9.79
  • cos(100exto)0.1736\text{cos}(100^ ext{o}) \approx -0.1736

Thus,

9.792=X2+1522X15(0.1736)9.79^2 = X^2 + 15^2 - 2 \cdot X \cdot 15 \cdot (-0.1736)

This simplifies to:

96.04=X2+225+5.208X96.04 = X^2 + 225 + 5.208 \cdot X

Rearranging gives us a quadratic equation:

X2+5.208X128.96=0X^2 + 5.208 X - 128.96 = 0

Using the quadratic formula:

X=b±b24ac2aX = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}

where,

  • a=1a = 1
  • b=5.208b = 5.208
  • c=128.96c = -128.96

Calculating the roots, we find:

X19.3NX \approx 19.3 \, \text{N}

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