Photo AI

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 1

Question icon

Question 1

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 1.png

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 1

Step 1

Find (a) the tension in the string.

96%

114 rated

Answer

To find the tension in the string, we'll use the equilibrium condition for forces acting on particle P, where the horizontal force (5 N) and the tension (T) interact.

Applying the condition for horizontal force:

5 = T imes rac{cos(25°)}{sin(65°)}

From Lami's Theorem, we can write:

T=5×sin(65°)sin(90°)T = \frac{5 \times sin(65°)}{sin(90°)}

Thus, simplifying gives us:

T=5×sin(65°)T = 5 \times sin(65°)

Calculating this:

T5×0.90634.5315 NT \approx 5 \times 0.9063 \approx 4.5315 \text{ N}

Therefore, the tension in the string is approximately 4.53 N.

Step 2

Find (b) the value of W.

99%

104 rated

Answer

To find the weight W of the particle P, we resolve the vertical forces.

Using the equilibrium condition for the vertical forces:

W=T×cos(25°)W = T \times cos(25°)

Substituting the value of T we calculated earlier:

W=(5×sin(65°))×cos(25°)W = (5 \times sin(65°)) \times cos(25°)

Calculating this:

First, we compute: sin(65°)0.9063sin(65°) \approx 0.9063 and cos(25°)0.9063cos(25°) \approx 0.9063

Thus:

W=(5×0.9063)×0.90634.53×0.90634.1151 NW = (5 \times 0.9063) \times 0.9063 \approx 4.53 \times 0.9063 \approx 4.1151 \text{ N}

Therefore, the value of W is approximately 4.12 N.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;