SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

  • Login
  • Get Help
  • About

ENGINEERING, PPT 2018 HSC 26c

The top view of a fidget spinner is shown.
 

What force Q is required to overcome the 20 N resistance force shown?   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`2.86\ text{N}`

Show Worked Solution

Actual diagram used to calculate the 35 mm distance below.

`sumM` `=0`  
`d` `=35\ text{(from scale drawing)}`  
`Q xx d` `=20 xx 5`  
`Q` `=100/35=2.857…`  

  
`:.\ Q = 2.86\ text{N}`


 
♦♦♦ Mean mark 27%.

Filed Under: Mechanics Tagged With: Band 6, smc-3718-85-Moments

ENGINEERING, PPT 2020 HSC 23b

A diagram of the front steering system of a bicycle is shown.
 


 

  1. A rider leans forward so that 55% of their mass is exerted as a force, as shown in the diagram. A turning moment of 132.19 Nm is generated about the front axle.
  2. What is the mass of the rider?   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

  1. In overcoming the resistance due to gravity, drag and rolling resistance, a rider applies a force of 500 N on the pedals. This allows the rider to cover, in 5 seconds, a distance of 20 metres from the starting point.
  2. How much power was produced by the rider?  (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

i.   `95\ text{kg}`

ii.  `2000\ text{W}`

Show Worked Solution

i.  
         

`sin20°` `= d/740`  
`d` `= 740 xx sin20°= 253.09\ text{mm}`  

 

`M` `= Fxxd`  
`F` `= 132.19/0.25309= 522.5\ text{N}`  

 

`55\text{%} xx F_(text{total})` `= 522.5\ text{N}`  
`F_(text{total})` `= 522.5/0.55= 950\ text{N}`  

 
`:. M_(text{rider})= 950/10= 95\ text{kg}`


♦ Mean mark 47%.

ii.   `F=500\ text{N}, \ s=20\ text{m},\ t=5\ text{sec}`

`P` `= Fxxv`  
  `= F xx s/t`  
  `= 500 xx 20/5`  
  `= 2000\ text{W}`  

Filed Under: Mechanics Tagged With: Band 4, Band 5, smc-3718-60-Work Energy Power, smc-3718-85-Moments

Copyright © 2014–2025 SmarterEd.com.au · Log in