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Mechanics, SPEC2 2019 VCAA 17 MC

A particle is held in equilibrium by three coplanar forces of magnitudes  `F_1, F_2`  and  `F_3`.

The angles between these forces are  `alpha, beta`  and  `gamma`  as shown in the diagram below.
 

If  `beta = 2alpha`, then  `(F_1)/(F_2)`  is equal to

  1. `1/2 sin(alpha)`
  2. `2sin(alpha)`
  3. `1/2text(cosec)(alpha)`
  4. `1/2cos(alpha)`
  5. `1/2sec(alpha)`
Show Answers Only

`E`

Show Worked Solution

`text(Using Lami’s theorem:)`

`(F_1)/(sin alpha)` `= (F_2)/(sinbeta)`
`(F_1)/(F_2)` `= (sin alpha)/(sin beta)`
  `= (sin alpha)/(sin 2alpha)`
  `= (sin alpha)/(2sin alphacos alpha)`
  `= 1/2 sec alpha`

 
`=>E`

Filed Under: Pulleys, Planes and Equilibrium (SM) Tagged With: Band 4, smc-1175-40-Equilibrium, smc-1175-50-Lami's theorem

Mechanics, SPEC2-NHT 2018 VCAA 14 MC

The diagram above shows a particle at `O` in equilibrium in a plane under the action of three forces of magnitudes `P, Q` and `R`.

Which one of the following statements is false?

A.   `R = Q sin (60^@)`

B.   `Q = R sin (60^@)`

C.   `P = R sin(30^@)`

D.   `Q cos (60^@) = P cos (30^@)`

E.   `P cos (60^@) + Q cos (30^@) = R`

Show Answers Only

`A`

Show Worked Solution

 

`text(By elimination:)`

`sin 60^@` `= Q/R`
`Q` `= R\ sin 60^@ \ \ text{(B correct)}`
`sin30^@` `= P/R`
`P` `= R\ sin30^@\ \ text{(C correct)}`

 

`text(Equating horizontal forces:)`

`Q\ cos 60^@ = P\ cos 30^@\ \ text{(D correct)}`

 

`text(Equating vertical forces:)`

`P\ cos 60^@ + Q\ cos 30^@ = R\ \ text{(E correct)}`

 
`=>  A`

Filed Under: Pulleys, Planes and Equilibrium (SM) Tagged With: Band 4, smc-1175-40-Equilibrium, smc-1175-50-Lami's theorem

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