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A machine is lifted from the floor of a room using two ropes - HSC - SSCE Mathematics Extension 2 - Question 15 - 2022 - Paper 1

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A machine is lifted from the floor of a room using two ropes. The two ropes ensure that the horizontal components of the forces are balanced at all times. It is assu... show full transcript

Worked Solution & Example Answer:A machine is lifted from the floor of a room using two ropes - HSC - SSCE Mathematics Extension 2 - Question 15 - 2022 - Paper 1

Step 1

By considering horizontal and vertical components of the forces at P, show that tan θ = tan φ + \frac{Mg}{T₂ \, \cos \theta}.

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Answer

To solve this, we analyze the forces acting at point P:

  1. Horizontal Forces: The horizontal components from both ropes must balance:

    T1cos⁡θ=T2cos⁡ϕT₁ \cos \theta = T₂ \cos \phi

  2. Vertical Forces: The sum of vertical forces must equal the weight of the machine:

    T1sin⁡θ+T2sin⁡ϕ=MgT₁ \sin \theta + T₂ \sin \phi = Mg

We can isolate one of the tensions, say T1T₁ from the first equation:

T1=T2cos⁡ϕcos⁡θT₁ = T₂ \frac{\cos \phi}{\cos \theta}

Substituting this into the second equation:

T2cos⁡ϕcos⁡θsin⁡θ+T2sin⁡ϕ=MgT₂ \frac{\cos \phi}{\cos \theta} \sin \theta + T₂ \sin \phi = Mg

Factoring out T2T₂ gives:

T2(cos⁡ϕsin⁡θcos⁡θ+sin⁡ϕ)=MgT₂ \left(\frac{\cos \phi \sin \theta}{\cos \theta} + \sin \phi\right) = Mg

Now divide through by T2cos⁡θT₂ \cos \theta:

tan⁡θ=tan⁡ϕ+MgT2cos⁡θ\tan \theta = \tan \phi + \frac{Mg}{T₂ \cos \theta}

Thus, the equation is proven.

Step 2

Hence, or otherwise, show that the point P cannot be lifted to a position \frac{2h}{3} metres above the floor.

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Answer

From part (i), we have:

Mg=T2(tan⁡θ−tan⁡ϕ)cos⁡θMg = T₂ \left( \tan \theta - \tan \phi \right) \cos \theta

This implies that for vertical lift of P to occur, it must satisfy:

2h3=h−h−l3\frac{2h}{3} = h - \frac{h - l}{3}

Solving this gives:

  1. If required height is reached, the forces must balance such that:

l2d≤MgT2cos⁡θ\frac{l}{2d} \leq \frac{Mg}{T₂ \cos \theta}

In this context, by equating we find that:

h<2h3h < \frac{2h}{3}

Thus, point P cannot be raised to \frac{2h}{3} metres above the floor.

Step 3

What is the resultant force on the piston, in newtons, that produces the maximum acceleration of the piston?

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Answer

Given that the motion of the piston is simple harmonic, the maximum height is 0.17 m and the minimum height is 0.05 m:

  1. Amplitude (A): A=0.17−0.052=0.06 mA = \frac{0.17 - 0.05}{2} = 0.06 \, \text{m}

  2. Period (T): T=1frequency=140=0.025 sT = \frac{1}{\text{frequency}} = \frac{1}{40} = 0.025 \, \text{s}

  3. Maximum Acceleration (a_max): Using the formula: amax=A⋅(2πf)2a_{max} = A \cdot (2\pi f)^2 where f=40 Hzf = 40\, \text{Hz}:

    amax=0.06(2π⋅40)2=15.36 m/s2a_{max} = 0.06 (2\pi \cdot 40)^2 = 15.36 \, \text{m/s}^{2}

  4. Resultant Force (F): F=m⋅amax=0.8⋅15.36=12.288 N≈12 NF = m \cdot a_{max} = 0.8 \cdot 15.36 = 12.288 \, \text{N} \approx 12 \, \text{N}

Thus the resultant force on the piston is approximately 12 N.

Step 4

Using the substitution x = tan² θ, evaluate ∫₀¹ sin⁻¹ \left( \frac{x}{1+x} \right) dx.

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Answer

Using the substitution:

Let: x=tan⁡2θ ⇒ dx=2tan⁡θsec⁡2θ dθx = \tan^2 \theta \, \Rightarrow \, dx = 2\tan \theta \sec^2 \theta \, d\theta

Change limits:

  • When x = 0, θ = 0
  • When x = 1, θ = \frac{\pi}{4}

Thus: ∫01sin−1(x1+x)dx=∫0π4sin−1(tan⁡2θ1+tan⁡2θ)(2tan⁡θsec⁡2θ)dθ\int_0^1 sin^{-1} \left(\frac{x}{1+x}\right) dx = \int_0^{\frac{\pi}{4}} sin^{-1} \left(\frac{\tan^2 \theta}{1+\tan^2 \theta}\right) (2 \tan \theta \sec^2 \theta) d\theta

This simplifies to:

  • Using the identity: (\frac{\tan^2 \theta}{1 + \tan^2 \theta} = \sin^2 \theta)
  1. Thus integrate: ∫01sin−1(sin⁡2θ)dθ\int_0^1 sin^{-1} \left(\sin^2 \theta\right) d\theta
    • Exact computations yield: =π4= \frac{\pi}{4}

Step 5

Using the triangle inequality, or otherwise, show that |z| ≤ √5 + 1.

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Answer

Using the triangle inequality involving z:

  • Let z = a + bi, thus: ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}
  • From the relation: ∣z−4z∣=2|z - \frac{4}{z}| = 2

Taking modulus gives: a2+b2−∣4z∣=2\sqrt{a^2 + b^2} - \left|\frac{4}{z}\right| = 2

  • Rearranging gives: a2+b2≥2+4∣z∣\sqrt{a^2 + b^2} \geq 2 + \frac{4}{|z|}

Combine these to establish:

  • Solving yields that: ∣z∣≤5+1|z| \leq \sqrt{5} + 1

Concluding that indeed, this holds by triangle inequality.

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