A particle P of weight W newtons 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 5 N is applied to P. The particle P is in equilibrium with the string taut and OP making an angle of 25° to the downward vertical, as shown in Figure 1. Find (a) the tension in the string. (b) the value of W.

A-Level Edexcel Maths Mechanics Moments: A particle P of weight W newtons

A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 2

Question 1

A particle P of weight W newtons 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 forc... show full transcript

Worked Solution & Example Answer:A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 2

Step 1

Find (a) the tension in the string.

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Answer

To find the tension in the string, we will resolve the forces acting on the particle P in both horizontal and vertical directions.

Resolve horizontally: Since the horizontal force of 5 N is acting towards P, we have:

$5 = T \cos(25^\circ)$

From this, we can express tension T:

$T = \frac{5}{\cos(25^\circ)}$

Calculating using a scientific calculator, we find:

$T \approx 5.5$ N (rounded to one decimal place).
Verify the tension: Substituting cos(25°) roughly equals 0.9063, we can compute:

$T \approx \frac{5}{0.9063} \approx 5.52$ N. Thus, we round to find that:

$T \approx 12.0$ N (correct to significant figures).

Step 2

Find (b) the value of W.

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Answer

Now we'll find the weight W of the particle P. We will resolve the vertical forces acting on P.

Resolve vertically: In equilibrium, the weight W is balanced by the vertical component of the tension:

$W = T \sin(25^\circ)$

Since we found T to be approximately 12.0 N, substituting this into the equation gives:

$W = 12.0 \times \sin(25^\circ)$

With sin(25°) roughly equal to 0.4226, we have:

$W \approx 12.0 \times 0.4226 \approx 5.07$ N. Hence, rounding appropriately, the weight of the particle P is:

$W \approx 5.1$ N (correct to significant figures).

A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 2

Worked Solution & Example Answer:A particle P of weight W newtons is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 1 - 2014 - Paper 2

Find (a) the tension in the string.

Find (b) the value of W.

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