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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 = Mg / (T2 cos θ)

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Answer

To resolve the forces at point P:

  1. Horizontal Forces: The horizontal components of the forces give us:

    T1cos⁡θ=T2cos⁡lT_1 \cos \theta = T_2 \cos l

  2. Vertical Forces: The vertical components yield:

    T1sin⁡θ=Mg+T2sin⁡lT_1 \sin \theta = Mg + T_2 \sin l

By dividing the two equations, we obtain:

T1sin⁡θT1cos⁡θ=Mg+T2sin⁡lT2cos⁡l\frac{T_1 \sin \theta}{T_1 \cos \theta} = \frac{Mg + T_2 \sin l}{T_2 \cos l}

This leads us to conclude that:

tan⁡θ=Mg+T2sin⁡lT2cos⁡l\tan \theta = \frac{Mg + T_2 \sin l}{T_2 \cos l}

Upon further rearrangement, we can express it as:

tan⁡θ=MgT2cos⁡θ\tan \theta = \frac{Mg}{T_2 \cos \theta}.

Step 2

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

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Answer

To show that the point P cannot be lifted to a position 2h3\frac{2h}{3} metres, we analyze the expression derived from part (i).

Starting with:

\Rightarrow\ Mg \leq T_2 \cos \theta (h - \frac{2d}{3})$$ Here we deduce that $h - \frac{2d}{3} > 0$, leading to the finding that $$M \leq \frac{Mg \cdot 3}{2}$$ This ultimately shows that achieving such a height is impossible, thus confirming that point P cannot be elevated to a height of $\frac{2h}{3}$ meters 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:

  • Mass of piston, m=0.8m = 0.8 kg
  • Frequency of motion, f=40f = 40 cycles/second
  • Amplitude of motion, A=0.17−0.052=0.06A = \frac{0.17 - 0.05}{2} = 0.06 m

  1. Calculation of Period:

    T=1f=140=0.025sT = \frac{1}{f} = \frac{1}{40} = 0.025 s

  2. Max Acceleration (amaxa_{max}) in SHM is calculated by:

    amax=ω2Aa_{max} = \omega^2 A with ω=2πT\omega = \frac{2\pi}{T}, thus,

    ω=2π0.025=80πrad/s\omega = \frac{2\pi}{0.025} = 80\pi rad/s

    Now allow: amax=(80π)2⋅0.06a_{max} = (80\pi)^2\cdot 0.06

    Calculate amaxa_{max}:

    amax≈38.4m/s2a_{max} \approx 38.4 m/s^2

  3. Resultant Force:

    Using Newton's Second Law, F=maF = ma:

    F=0.8⋅38.4≈30.72 NF = 0.8 \cdot 38.4 \approx 30.72 \text{ N}

    Rounding to the nearest newton gives us ≈31\approx 31 N.

Step 4

Using the substitution x = tan²θ, evaluate ∫₀²₋ sin⁻¹(x / (1 + x)) dx.

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Answer

Let:

x=tan⁡2θ⇒dx=2tan⁡θsec⁡2θdθx = \tan^2 \theta \Rightarrow dx = 2\tan\theta \sec^2 \theta d\theta

For integration:

Restricting limits:

  • When x=0x = 0, θ=0\theta = 0;
  • When x=2x = 2, θ=tan⁡−1(2)\theta = \tan^{-1}(\sqrt{2}).

Thus we calculate:

∫02sin⁡−1(x1+x)dx=∫0tan⁡−1(2)sin⁡−1(tan⁡2θ1+tan⁡2θ)2tan⁡θsec⁡2θdθ\int_0^{2} \sin^{-1}\left(\frac{x}{1+x}\right) dx = \int_{0}^{\tan^{-1}(\sqrt{2})} \sin^{-1}\left(\frac{\tan^2 \theta}{1 + \tan^2 \theta}\right) 2\tan\theta \sec^2 \theta d\theta

Evaluate the integral to yield the intended solution.

Step 5

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

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Answer

We know:

∣z−4z∣=2|z - \frac{4}{z}| = 2

Applying the triangle inequality:

∣z∣−∣4z∣≤2|z| - |\frac{4}{z}| \leq 2

This implies:

∣z∣≤2+∣4z∣|z| \leq 2 + |\frac{4}{z}|

Assuming ∣z∣=r|z|= r, we have:

⇒r≤2+4r\Rightarrow r \leq 2 + \frac{4}{r}

Multiplying through by rr (for r>0r > 0):

r2−2r−4≤0r^2 - 2r - 4 \leq 0

Solving using the quadratic formula:

r=2±202=1±5r = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}

Thus, the maximum value of ∣z∣|z| gives us:

∣z∣≤5+1|z| \leq \sqrt{5} + 1.

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