Photo AI
A particle of mass 24 kg is attached to two light elastic strings, each of natural length 33 cm and elastic constant k - Leaving Cert Applied Maths - Question 7 - 2011
Question 7
A particle of mass 24 kg is attached to two light elastic strings, each of natural length 33 cm and elastic constant k. The other ends of the strings are attached t... show full transcript
Worked Solution & Example Answer:A particle of mass 24 kg is attached to two light elastic strings, each of natural length 33 cm and elastic constant k - Leaving Cert Applied Maths - Question 7 - 2011
Step 1
Show that the extension of each string is 7 cm.
Answer
To show that the extension of each string is 7 cm, we start by analyzing the geometry of the situation.
Using the trigonometric relationships,
-
We can find the length of the strings when the mass is hanging:
- The angle α can be found using: [ \tan(\alpha) = \frac{3}{4} \rightarrow \alpha = \tan^{-1}\left(\frac{3}{4}\right) ]
-
Using the cosine rule in the triangle formed, we can relate the lengths:
- For the horizontal distance, we have:
[ 64 = 2 \cdot (33 + x) \cdot \cos(\alpha) ]
- From the triangle properties:
- Let ( cos(\alpha) =\frac{4}{5}), then we can set up the equation: [ \frac{4}{5} = \frac{32}{33 + x} ]
- For the horizontal distance, we have:
[ 64 = 2 \cdot (33 + x) \cdot \cos(\alpha) ]
-
Rearranging gives us: [ 32 = \frac{4}{5}(33 + x) ] [ 32 \cdot 5 = 4(33 + x) ] [ 160 = 132 + 4x ] [ 4x = 160 - 132 = 28 ] [ x = 7 \mathrm{~cm} ]
- Thus, the extension of each string is confirmed to be 7 cm.
Step 2
Find the value of k.
Answer
To find the value of k, we analyze the forces acting on the mass:
-
The vertical forces include the weight of the mass and the tension in the strings:
- The weight (W) is given by: [ W = 24g ]
-
The tension in the strings can be expressed using the angle made:
- For the vertical component of the tension, we have: [ 2T \cdot \sin(\alpha) = 24g ]
-
Substituting for ( \sin(\alpha) ):
- Using ( \tan(\alpha) = \frac{3}{4} \rightarrow \sin(\alpha) = \frac{3}{5} ):
- We find: [ 2T \cdot \frac{3}{5} = 24g ]
- Rearranging gives: [ T = \frac{20g}{3} ]
-
The extension in the string relates to the spring constant k as: [ T = kx \rightarrow T = k(0.07) ]
-
Thus we get: [ \frac{20g}{3} = k(0.07) ]
- Solving gives: [ k = \frac{20g}{3(0.07)} ]
- On substituting g = 10 m/s²:
[ k = \frac{20 \cdot 10}{3 \cdot 0.07} = 2800 \mathrm{
Nm^{-1}} ]
Therefore, the value of k is 2800 N/m.
Step 3
Prove that µ ≥ tan θ.
Answer
To prove that µ ≥ tan θ:
-
Starting from the equilibrium conditions of the rod:
- The tension T in the string and the frictional forces acting on point B help define the equation: [ T\sin(2θ) = W\sin(2ρ) ]
-
Replacing ( T ) with its expression: [ T = W \cos(θ) ]
-
Using the resulting equations: [ µR + T \cos(θ) = W ]
-
Breaking the forces down, we see:
- Applying the friction coefficient: [ µT \sin(θ) + T \cos(θ) = W ]
-
Simplifying gives:
- Conclusively: [ µ = \frac{T \sin(θ)}{T \cos(θ)} = \tan(θ) \rightarrow µ ≥ tan(θ) ]
Thus, we have shown that µ ≥ tan θ.