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ENGINEERING, PPT 2025 HSC 23c

A winch is a gear system. It can be used to pull a boat out of the water and up a slipway.

A 1.2-tonne boat inclined at 30 degrees to the horizontal plane is being pulled up a slipway by a winch, as shown.
 
 
         

  1. The coefficient of friction between the boat and the slipway is 0.4 and the winch cable is positioned 40 degrees below the horizontal.
  1. Calculate the minimum force required in the winch cable to pull the boat up the slipway.   (5 marks)

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  2. If the velocity ratio (VR) of the system is 2, calculate the efficiency.   (3 marks)

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i.    \(9633.87\ \text{N}\)

ii.   \(62.3\%\)

Show Worked Solution

i.    Trigonometric solution

\(W\) \(=1200\times10=12\ 000\ \text{N}\)
\(\phi\) \(=\tan^{-1}(0.4)=21.8^{\circ}\)

    
The resultant \(R\) acts at \((30^{\circ}+21.8^{\circ})=51.8^{\circ}\) from the vertical.

Angle between \(T\) and vertical \(=180^{\circ}-50^{\circ}-51.8^{\circ}=78.2^{\circ}\)

 

\(\dfrac{T}{\sin 51.8^{\circ}}\) \(=\dfrac{12\ 000}{\sin 78.2^{\circ}}\)  
\(T\) \(=\dfrac{12\ 000\times\sin 51.8^{\circ}}{\sin 78.2^{\circ}}=9633.87\ \text{N}\)  

 

Alternative graphical solution

\(\mu=0.4,\ \ \phi=\tan^{-1}0.4=21.8^{\circ}\)

\(\text{Scale }1\ \text{mm}=100\ \text{N}\)

\(W\) \(=12\ 000\ \text{N}\rightarrow120\ \text{mm}\)
\(T\) \(=100\ \text{mm}\rightarrow T\approx10\ \text{kN}\ (9633.87\ \text{N})\)

 

ii.    \(MA\) \(=\dfrac{L}{E}=\dfrac{12\ 000}{9633.87}=1.2456\)
  \(\eta\) \(=\dfrac{MA}{VR}\times100\%=\dfrac{1.2456}{2}\times100\%=62.3\%\)

Filed Under: Mechanics Tagged With: Band 4, smc-3718-10-Mechanical advantage, smc-3718-25-Efficiency, smc-3718-30-Friction, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2024 HSC 10 MC

An inclined plane is 5 metres long and 2 metres high. A force of 50 N is required to push a box up along the length of the slope.
 

Assuming no friction and 100% efficiency, what is the mass of the box to the nearest kilogram?

  1. 5
  2. 13
  3. 20
  4. 125
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\(B\)

Show Worked Solution

\(\text{Energy moving box}\ = 50 \times 5 = 250\ \text{J}\)

\(\text{Energy used = increase in potential energy of box (100% efficiency)}\)

\(m \times g \times h\) \(=250\)  
\(m\) \(=\dfrac{250}{10 \times 2}=12.5\ \text{kg}\)  

 
\(\Rightarrow B\)

♦ Mean mark 40%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-50-Inclined planes

ENGINEERING, PPT 2023 HSC 18 MC

A diagram of a block being pulled by a rope up an inclined plane is shown.
 

Which free body diagram correctly represents the diagram given?
 


 

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\(  C \)

Show Worked Solution

\(\Rightarrow  C \)

Filed Under: Mechanics Tagged With: Band 4, smc-3718-50-Inclined planes

ENGINEERING, PPT 2017 HSC 3 MC

A mass `M` sits on a plane inclined at `theta`° to the horizontal.
 

What expression is used to calculate the normal reaction `N` ?

  1. `Mg`
  2. `Mg\ costheta`
  3. `Mg\ sintheta`
  4. `Mg\ tantheta`
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`B`

Show Worked Solution

`=>B`

Filed Under: Mechanics Tagged With: Band 4, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2018 HSC 20 MC

A box with weight `W` and subject to a friction force `F` is being pulled up an inclined plane by a force `P`.
 

The force `R` is the resultant of which two forces?

  1. `P` and `W`
  2. `P` and `N`
  3. `F` and `W`
  4. `F` and `N`
Show Answers Only

`D`

Show Worked Solution
  • `R` is used as the resultant of `F` and `N`, usually used to calculate `P` and employ the 3 force rule.

`=>D`


♦ Mean mark 42%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-30-Friction, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2017 HSC 24b

A child and sled with a combined mass of 23 kg are being pulled along a horizontal snow-covered surface using a rope.
 

The coefficient of static friction between the sled and the snow is 0.14.

  1. Draw a free-body diagram that indicates the forces acting on the sled. Label the diagram.   (2 marks)

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  1. Calculate the tension in the rope immediately before the point at which the sled and child begin to move.   (3 marks)

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i.
       

 

ii.   `text{32 N}`

Show Worked Solution
i.
       

Mean mark (i) 57%.

ii.   `text{Weight}\ = mg = 23 xx 10 = 230\ text{N}`     

`text{Solving for}\ T:`

`tan^(-1)(0.14) = 8°`

`text{Using the sine rule:}`

`T/(sin8°)` `=230/(sin92°)`  
`T` `=(230 xx sin8°)/(sin92°)=32.0\ text{N}`  

  
`:.\ text{Tension in the rope = 32 N}`


♦ Mean mark (ii) 40%.

Filed Under: Mechanics Tagged With: Band 4, Band 5, smc-3718-30-Friction, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2020 HSC 14 MC

The diagram shows the forces acting on a block on an inclined plane. The block is held in a stationary position by force `P`.
 

What is the correct equation for the normal force, `N` ?

  1. `N=W cos theta - P sin phi`
  2. `N=W cos theta + P sin phi`
  3. `N=W sin theta - P cos phi`
  4. `N=W sin theta + P cos phi`
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`A`

Show Worked Solution

  • `P\ sin phi` is the upward force on the block perpendicular to the plane.

  • `Wcos\ theta`  is the downward force on the block perpendicular to the plane.
  • Therefore, `N` is equal to the downward component of the weight force minus the upwards component of force `P`.

`=>A`


♦ Mean mark 50%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-40-Normal Force, smc-3718-50-Inclined planes

ENGINEERING, PPT 2021 HSC 26b

The diagram shows a box which is at rest on an inclined ramp. The length of the ramp is 3 m and one end is raised `h` metres off the ground.
 


 

The coefficient of friction between the ramp and the box is 0.5.

To what height `(h)` must the ramp be raised for the box to just begin to slide?   (3 marks)

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`1.34\ text{m}`

Show Worked Solution
`mu` `= 0.5 = tan\ theta`  
`theta`  `= tan^(-1)(0.5)=  26.57°`  

 

`sin(26.57°)`  `= h/3`  
`:.h` ` = 3 xx sin(26.57°)= 1.34\ text{m}`  

Filed Under: Mechanics Tagged With: Band 4, smc-3718-30-Friction, smc-3718-50-Inclined planes

ENGINEERING, PPT 2021 HSC 22d

A skateboard rider is at rest at the top of a slope, indicated by point `A` in the diagram.
 

The combined mass of the rider and skateboard is 80 kg.

Calculate the speed of the rider, in metres per second, at point `B` if the average work done by friction against the skateboard is 55 N.    (4 marks)

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`text{Speed} = 14.57\ text{m/s}`

Show Worked Solution

`m= 80\ text{kg}, \ f= 55\ text{N}`

`PE= mgh= 80xx10xx27.8= 22\ 240\ text{J}`

   
`text{Losses to friction (W):}`

`W= fs= 55xx250= 13\ 750\ text{J}`
 

`text{Energy used}` `= PE−text{Frictional losses}`  
  `= 22\ 240−13\ 750`  
  `= 8490\ text{J}`  

 
`text{Calculate speed at point B:}`

`1/2×m×v^2` `= 8490`  
`0.5×80×v^2` `= 8490`  
`v^2` `= 212.25`  
`:.v` `= 14.57\ text{m/s}`  

♦ Mean mark 43%.

Filed Under: Mechanics Tagged With: Band 5, smc-3718-30-Friction, smc-3718-50-Inclined planes, smc-3718-60-Work Energy Power

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