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Aussie Maths & Science Teachers: Save your time with SmarterEd

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}`  

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`

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→ `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%.

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)
  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}`

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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%.

ENGINEERING, PPT 2022 HSC 4 MC

The diagram shows a box being pulled along a horizontal surface at angle `theta` to the horizontal.
 

Which equation is correct for `N`?

  1. `N = W - P costheta`
  2. `N = W + P costheta`
  3. `N = W - P sintheta`
  4. `N = W + P sintheta`
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`C`

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`SigmaF_V` `=0`  
`0` `= N-W + P sintheta`  
`N` `= W-P sintheta`  

 
`=>C`

Mean mark 59%.

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`

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→ `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%.