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A particle P is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2007 - Paper 1

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A particle P is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. A horizontal force of magnitude 12 N ... show full transcript

Worked Solution & Example Answer:A particle P is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2007 - Paper 1

Step 1

Find (a) the tension in the string.

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Answer

To find the tension in the string, we can use the equilibrium condition in the vertical direction. The vertical component of the tension must balance the horizontal force:

We know that: Textsin20°=12T ext{ sin } 20° = 12

Solving for T: T = rac{12}{ ext{sin } 20°}

Calculating:

  • First, find ext{sin } 20°: extsin20°extisapproximately0.342. ext{sin } 20° ext{ is approximately } 0.342.

Then: T ext{ is approximately } rac{12}{0.342} ext{ N} ext{ or } 35.1 ext{ N} ext{ (to one decimal place).}

Step 2

Find (b) the weight of P.

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Answer

To find the weight of P, we'll use the equilibrium conditions in the vertical direction again, where the weight is balanced by the vertical component of tension:

From the earlier tension calculation, we have:

  • Weight (W) is given by: W=Textcos20°W = T ext{ cos } 20°

Substituting in the tension value: W=35.1extNimesextcos20°W = 35.1 ext{ N} imes ext{cos } 20°

Calculating:

  • First, find ext{cos } 20°: extcos20°extisapproximately0.940. ext{cos } 20° ext{ is approximately } 0.940.

Then: Wextisapproximately35.1imes0.940extNextor33.0extNext(toonedecimalplace).W ext{ is approximately } 35.1 imes 0.940 ext{ N} ext{ or } 33.0 ext{ N} ext{ (to one decimal place).}

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