Photo AI
A smooth bead B is threaded on a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1
Question 3
A smooth bead B is threaded on a light inextensible string. The ends of the string are attached to two fixed points A and C on the same horizontal level. The bead is... show full transcript
Worked Solution & Example Answer:A smooth bead B is threaded on a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1
Step 1
(a) the tension in the string.
Answer
To find the tension in the string, we can analyze the forces acting on the bead B. Since the bead is in equilibrium, we can apply the following horizontal force balance:
[ T \cos \alpha = 6 , \text{N} ]
Rearranging gives:
[ T = \frac{6}{\cos \alpha} ]
Next, we need to determine ( \cos \alpha ) using the given ( \tan \alpha = \frac{2}{3} ). From the tangent function, we can find the adjacent side and hypotenuse in a right triangle:
[ \tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{3} ]
Using the Pythagorean theorem:
[ \text{hypotenuse} = \sqrt{3^2 + 2^2} = \sqrt{13} ]
Thus,
[ \cos \alpha = \frac{3}{\sqrt{13}} ]
Now substitute into the tension equation:
[ T = \frac{6}{\frac{3}{\sqrt{13}}} = \frac{6 \sqrt{13}}{3} = 2 \sqrt{13} , \text{N} \approx 7.5 , \text{N} ]
Therefore, the tension in the string is approximately 7.5 N.
Step 2
(b) the weight of the bead.
Answer
To find the weight of the bead, we can use the vertical force balance:
[ T \sin \alpha = W ]
Substituting for ( T ) from part (a):
[ W = T \sin \alpha ]
Next, we need to estimate ( \sin \alpha ) using the value of ( \tan \alpha = \frac{2}{3} ) again:
Using the triangle sides from part (a), we have:
[ \sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{13}} ]
Now, substitute:
[ W = 2 \sqrt{13} \cdot \frac{2}{\sqrt{13}} = 4 , \text{N} ]
Thus, the weight of the bead is 12 N.