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ENGINEERING, PPT 2024 HSC 19 MC

Which is the correct angle of repose for rock fill, used on a highway construction, that has a coefficient of friction of 0.84?

  1. 20°
  2. 30°
  3. 40°
  4. 50°
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\(C\)

Show Worked Solution

\(\mu = 0.84\)

\(\tan \theta\) \(=0.84\)  
\(\theta\) \(= \tan^{-1} 0.84=40^{\circ}\)  

 
\(\Rightarrow C\)

Filed Under: Mechanics Tagged With: Band 3, smc-3718-30-Friction

ENGINEERING, PPT 2023 HSC 25d

A uniform 8-metre ladder with a mass of 12 kg has been placed against a smooth wall. 
 

Determine the minimum coefficient of static friction between the ground and the ladder. Assume there is no friction between the ladder and the wall.   (4 marks)

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\(\mu = 0.287\)

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Graphical solution:

\(\phi = 16^{\circ}\)

\(\mu = \tan \phi = \tan 16^{\circ} = 0.287\ \text{(3 d.p.)}\)

♦♦ Mean mark 31%.

Analytical solution:

\(\cos 60^{\circ} = \dfrac{\text{d}_1}{4}\ \Rightarrow \ \text{d}_1 = 2\ \text{m} \)

\(\sin 60^{\circ} = \dfrac{\text{d}_2}{8}\ \Rightarrow \ \text{d}_2 = 6.928\ \text{m} \)

\( \Sigma \text{M}_{\text{G}} \) \(=0\)  
\(0\) \(=(120 \times 2)-(\text{F}_{\text{W}} \times 6.928) \)  
\(\text{F}_{\text{W}}\) \(=\dfrac{240}{6.928}= 34.64\ \text{N}\)  

 
\(\therefore \text{F}_{\text{F}} = 34.64 \text{ N}\)

\(\text{F}_{\text{g}}=mg=120\ \text{N}\)

 
\(\therefore \text{N} = 120 \text{N}\uparrow \)
 

\(\text{F}_{\text{F}}\) \(= \mu \text{N}\)  
\(\mu\) \(= \dfrac{34.64}{120}=0.289\ \text{(3 d.p.)} \)  

Filed Under: Mechanics Tagged With: Band 5, Band 6, smc-3718-30-Friction

ENGINEERING, PPT 2018 HSC 26b

The body of the fidget spinner is made from polypropylene. The bearing is made from steel.

When the force P applied by the press punch is 16 N, the bearing slides into position.
 

Calculate the normal force (N) acting on the bearing walls if the coefficient of friction between polypropylene and steel is 0.20.   (2 marks)

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`80\ text{N}`

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`F=16\ text{N},\ \ mu=0.2`

`F` `=muN`  
`N` `=F/mu=16/0.2=80\ text{N}`  

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

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`
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`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 2019 HSC 7 MC

A box sits on a horizontal surface. The box begins to slip when this surface is tilted to 28 degrees.

What is the coefficient of friction between the box and the surface?

  1. 0.280
  2. 0.469
  3. 0.532
  4. 0.883
Show Answers Only

`C`

Show Worked Solution

`mu = tan(phi) = tan28° = 0.532`

`=>C`

Filed Under: Mechanics Tagged With: Band 3, smc-3718-30-Friction

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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