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Mechanics, EXT2 2019 SPEC1 9

  1. A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.
     

  1. Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`.   (1 mark)

  2. A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.
     

  1. Show that  `F = mg((1-cos(beta))/(sin(beta)))`.   (3 marks)

Show Answers Only

a.   `T = (mg)/(2sinalpha)`

b.    `text(See Worked Solutions)`

Show Worked Solution
a.   
`2 xx Tsinalpha` `= mg`
`:.T` `= (mg)/(2sinalpha)`

 

b.   

`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)` `= mg`
`T` `= (mg)/(sin(beta) + sin(2beta))`

`text(Resolving forces horizontally:)`

`F + Tcos(2beta)` `= Tcos(beta)`
`F` `= Tcos(beta)-Tcos(2beta)`
  `= T(cos(beta)-cos(2beta))`
  `= T[cos(beta)-(2cos^2beta-1)]`
  `= T(−2cos^2(beta) + cos(beta) + 1)`
  `= T(−2cos(beta)-1)(cos(beta)-1)`
  `= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
  `= (mg(1-cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
  `= mg((1-cos(beta))/(sin(beta)))`

Mechanics, EXT2 EQ-Bank 30

A mass is acted upon by three forces (as shown in the diagram below). Express the resultant force using the standard unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) in the horizontal and vertical directions, and hence calculate the size of the resultant force acting on the mass. Describe the direction of this resultant force.   (4 marks)

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Show Worked Solution

\(\text{Resolving forces into \(\mathbf{i}\) and \(\mathbf{j}\) components}\)

\(\text{Resultant}\)

\(\text{Force:}\)

\(F\) \(=(-20+15 \cos 45+10 \cos 60) \mathbf{i}+(15 \sin 45-10 \sin 60)\mathbf{j}\)
  \(=\left(\dfrac{15 \sqrt{2}}{2}-15\right)\mathbf{i}+\left(\dfrac{15 \sqrt{2}}{2}-5 \sqrt{3}\right)\mathbf{j}\)
\(\abs{F}\) \(=\sqrt{\left(\dfrac{3 \sqrt{2}}{2}-15\right)^2+\left(\dfrac{5 \sqrt{2}}{2}-5 \sqrt{3}\right)^2} =4.805 \cdots=4.81 \ N\)

 

\(\text{Direction}\):

\(\tan \theta\) \(=\dfrac{\frac{15 \sqrt{2}}{2}-5 \sqrt{3}}{\frac{15 \sqrt{2}}{2}-15}=-0.443 \ldots\)
\(\theta\) \(=180-23.89=156^{\circ}\)

 

\(\text{i.e.} \ 156^{\circ} \ \text{direction measured from the \(x\)-axis}\)

Mechanics, EXT2 EQ-Bank 22

A 12 kilogram object is suspended from a horizontal ceiling by two light, inextensible strings at angles of 30° and 45°, as shown in the diagram.
 

The acceleration due to gravity is \(g\) m s\(^{-2}\) and the tensions in the strings are \(T_1\) newtons and \(T_2\) newtons.

  1. Show that \(T_2=\sqrt{\dfrac{3}{2}} T_1\)   (2 marks)

  2. Determine the tensions, in newtons, of \(T_1\) and \(T_2.\)   (2 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \(T_1=12(\sqrt{3}-1)g\)

\(T_2=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))\)

Show Worked Solution

a.    \(\text{Resolve forces into horizontal / vertical components:}\)
 

         

\(\text{Horizontal forces are equal.}\)

\(T_1\cos 30^{\circ}\) \(=T_2 \cos 45^{\circ}\)
\(T_1 \times \dfrac{\sqrt{3}}{2}\) \(=T_2 \times \dfrac{1}{\sqrt{2}}\)
\(T_2\) \(=\dfrac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \, T_1=\sqrt{\dfrac{3}{2}}\, T_1\)

 

b.    \(\text{Vertical forces are equal.}\)

\(T_1 \sin 30+T_2 \sin 45\) \(=12 g\)  
\(T_1 \times \dfrac{1}{2}+T_2 \times \dfrac{1}{\sqrt{2}}\) \(=12 g\)  

 
\(\text{Substitute} \ \ T_2=\sqrt{\dfrac{3}{2}}\, T_1:\)

\(T_1 \times \dfrac{1}{2}+T_1 \times \sqrt{\dfrac{3}{2}} \times \dfrac{1}{\sqrt{2}}\) \(=12 g\)
\(T_1\left(\dfrac{\sqrt{3}+1}{2}\right)\) \(=12 g\)

 
\(T_1=\dfrac{24g}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}=12(\sqrt{3}-1)g\)
 

\(T_2\) \(=\sqrt{\dfrac{3}{2}} \times 12(\sqrt{3}-1)g\)
  \(=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))\)

Mechanics, EXT2 EQ-Bank 17

Two light inextensible strings are attached to a horizontal surface and suspended a 10-kilogram object as shown in the diagram below 
 

The tension in the strings are \(T_1\) newtons and \(T_2\) newtons.

  1. Express \(T_1\) in terms of \(T_2\).   (2 marks)

  2. If the acceleration due to gravity is 9.8 ms\(^{-2}\), determine the exact values of \(T_1\) and \(T_2\), in newtons.   (2 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \(T_1=49 \sqrt{3}, T_2=49\)

Show Worked Solution

a.    \(\text{Resolve forces into horizontal/vertical components:}\)

 
       

\(\text{Horizontal forces are equal.}\)

\(T_1 \cos 60^{\circ}\) \(=T_2 \cos 30^{\circ}\)
\(T_1 \times \dfrac{1}{2}\) \(=T_2 \times \dfrac{\sqrt{3}}{2}\)
\(T_1\) \(=\sqrt{3}\, T_2\)

 

b.    \(\text{Vertical forces are equal.}\)

\(T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}\) \(=10 \times 9.8\)
\(T_1 \times \dfrac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}\) \(=98\)

 

\(\text {Substitute} \ \ T_1=\sqrt{3}\, T_2 :\)

\(\sqrt{3}\, T_2 \times \dfrac{\sqrt{3}}{2}+T_2 \times \dfrac{1}{2}\) \(=98\)
\(2\, T_2\) \(=98\)
\(T_2\) \(=49 \ \text{newtons}\)

 

\(\therefore T_1=49 \sqrt{3}, \ T_2=49\)

Mechanics, EXT2 EQ-Bank 24

A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.
 

  1. Find the normal reaction force, in newtons, that the surface exerts on the mass.   (2 marks)
  2. Find the acceleration of the mass, in ms−2, after it begins to move.   (2 marks)
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a.    `R = 2g-5/2-5 sqrt 3\ text(N)`

b.    `ddot{x} = 5/2-(5 sqrt 3)/2\ text(ms)^(-2)`

Show Worked Solution
a.  

`text{Let}\ R =\ text{normal force}`

`text(Resolving forces vertically:)`

`2g` `= 5 sin 30 + 10 sin 60 + R`
`2g` `= 5/2 + 5 sqrt 3 + R`
`R` `= 2g-5/2-5 sqrt 3\ text(N)`

 
b.    `text{Using}\ Sigma F=m ddotx :`

  `2ddot{x}` `= 10 cos 60-5 cos 30`
  `2ddot{x}` `= 5-(5 sqrt 3)/2`
  `:.ddot{x}` `= 5/2-(5 sqrt 3)/4\ text(ms)^(-2)`

Mechanics, EXT2 2020 SPEC2 18 MC

A particle of mass `m` kilogram hangs from a string that is attached to a fixed point. The particle is acted on by a horizontal force of magnitude `F` newtons. The system is in equilibrium when the string makes an angle `alpha` to the horizontal, as shown in the diagram below. The tension in the string has magnitude `T` newtons.
 

The value of  `tan\ alpha`  is

  1. `(mg)/T`
  2. `T/(mg)`
  3. `F/(mg)`
  4. `(mg)/F`
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`D`

Show Worked Solution

`text(Resolving forces vertically:)`

`mg = Tsin(alpha)`
 

`text(Resolving forces horizontally:)`

`F = Tcos(alpha)`

`:. tan\ alpha = (mg)/F`

`=>D`

Mechanics, EXT2 M1 2022 HSC 15a

A machine is lifted from the floor of a room using two ropes. The two ropes ensure that the horizontal components of the forces are balanced at all times. It is assumed that at all times the machine moves vertically upwards at a constant velocity.

The machine is located in a room with height `h` metres.

One of the ropes is attached to the point `P` on the machine and to the fixed point `C` on the ceiling of the room. The point `C` is a distance `d` metres to the left of `P`. Let the vertical distance from `P` to the ceiling be `ℓ` metres and let `\theta` be the angle this rope makes with the horizontal.

The other rope is attached to the point `P` and to the fixed point `F` on the floor of the room. The point `F` is a distance `2 d` metres to the right of `P`. Let `\phi` be the angle this rope makes with the horizontal.

Let the tension in the first rope be `T_1` newtons, the tension in the second rope be `T_2` newtons, the mass of the machine be `M` kilograms and the acceleration due to gravity be `g\ text{m s}^(-2)`.
 

  1. By considering horizontal and vertical components of the forces at `P`, show that
  2.            `tan theta=tan phi+(Mg)/(T_(2)cos phi)`   (3 marks)

  3. Hence, or otherwise, show that the point `P` cannot be lifted to a position `{2 h}/{3}` metres above the floor.   (2 marks)

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i.    `text{Proof (See Worked Solutions)}`

ii.   `text{Proof (See Worked Solutions)}`

Show Worked Solution

i.   
       

`text{Resolving forces horizontally:}`

`T_1cos\ theta=T_2cos\ phi\ \ text{… (1)}`

`text{Resolving forces vertically:}`

`T_1\ sin\ theta=Mg + T_2\ sin\ phi\ \ text{… (2)}`
 

`text{Divide}\ (2) -: (1):`

`tan\ theta=(Mg + T_2\ sin\ phi)/(cos\ phi)=(Mg)/(T_2\ cos\ phi)+tan\ phi`
 


Mean mark (i) 56%.

ii.   `text{Using diagram in part (i):}`

`tan\ theta=l/d,\ \ tan\ phi=(h-l)/(2d)`

`text{Using part (i):}`

`l/d=(Mg)/(T_2\ cos\ phi)+tan\ phi=(Mg)/(T_2\ cos\ phi)+(h-l)/(2d)`
 

`text{S}text{ince}\ \ (Mg)/(T_2\ cos\ phi)>0`

`l/d` `>(h-l)/(2d)`  
`l` `>(h-l)/2`  
`(3l)/2` `>h/2`  
`l` `>h/3`  
`h-l` `>h-h/3`  
`h-l` `>(2h)/3`  

 
`:.\ text{Point}\ P\ text(cannot be lifted to)\ (2h)/3\ text(metres above floor.)`


♦♦♦ Mean mark (ii) 27%.

Mechanics, EXT2 M1 2021 SPEC2 15

The diagram below shows a stationary body being acted on by four forces whose magnitudes are in newtons. The force of magnitude `F_1` newtons acts in the opposite direction to the force of magnitude 8 N.
 

Calculate the value of `F_1` in newtons.   (3 marks)

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`8-2sqrt3\ \ text(N)`

Show Worked Solution

`text(Resolve forces vertically:)`

`F_2sin60^@` `= 6sin30^@`
`F_2 sqrt3/2` `= 6 xx 1/2`
`F_2` `= 6/sqrt3= 2sqrt3`

 
`text(Resolve forces horizontally:)`

`F_1 + 6cos30^@` `= 2sqrt3 cos60^@ + 8`
`F_1 + 3sqrt3` `= sqrt3 + 8`
`F_1` `= 8-2sqrt3\ \ text(N)`

Mechanics, EXT2 M1 2021 HSC 14b

An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is `g\ text(m s)^(-2)` and the velocity of the object down the slope is `v\ text(m s)^(-1)`.

As well as the force due to gravity, the object is acted on by two forces, one of magnitude `2v` newtons and one of magnitude `2v^2` newtons, both acting up the slope.

  1. Show that the resultant force down the slope is  `(5sqrt3)/2 g-2v-2v^2`  newtons.   (2 marks)

  2. There is one value of `v` such that the object will slide down the slope at a constant speed.
  3. Find this value of `v` in `text(m s)^(-1)`, correct to 1 decimal place, given that  `g = 10`.   (2 marks)

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i.    `text(See Worked Solutions)`

ii.   `4.2\ \ text(ms)^-1`

Show Worked Solution

i.   

`F_s` `=\ text(force down slope)`
  `= 5g cos30-2v^2-2v`
  `= 5g · sqrt3/2-2v^2-2v`
  `= (5sqrt3)/2 g-2v^2-2v`

 

ii.   `text(Constant speed occurs if)\ \ a = 0`

`F_s = ma = 0`

`2v^2 + 2v-(5sqrt3)/2 xx 10` `= 0`
`2v^2 + 2v-25sqrt3` `= 0`

 

`v` `= (-2 + sqrt(2^2 + 4 · 2 · 25sqrt3))/(2 xx 2)`
  `= (-2 + sqrt(4 + 200sqrt3))/4`
  `= (-1 + sqrt(1 + 50 sqrt3))/2`
  `= 4.179…`
  `= 4.2\ \ text(ms)^-1\ text{(1 d.p.)}`

Mechanics, EXT2 M1 2020 HSC 12a

A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a resistive force of `0.3R` newtons, where `R` is the normal force, as shown in the diagram.

Take the acceleration `g` due to gravity to be 10m/s2.
 

  1. By resolving the forces vertically, show that  `R =400`.   (2 marks)

  2. Show that the net force horizontally is approximately 53.2 newtons.   (2 marks)

  3. Find the velocity of the box after the first three seconds.   (2 marks)

Show Answers Only

i.    `text{See Worked Solutions}`

ii.   `text{See Worked Solutions}`

iii.  `3.19 \ text{ms}^-1`

Show Worked Solution

i.   

`text{Resolving forces vertically:}`

`R + 200 \ sin 30^@` `= 50g`
`R + 200 xx frac{1}{2}` `= 50 xx 10`
`R + 100` `= 500`
`therefore \ R` `= 400 \ text(N)`

 
ii.    `text{Resolving forces horizontally:}`

`text{Net Force}` `= 200 \ cos 30^@-0.3 R`
  `= 200 xx frac{sqrt3}{2}-0.3 xx 400`
  `= 100 sqrt3-120`
  `= 53.2 \ text{N (to 1 d. p.)}`

 

iii.    `F` `=ma`
  `50 a` `=100 sqrt300-120`
  `a` `= frac{100 sqrt3-120}{50}\ text(ms)^(-2)`

  
`text{Initially,}\ u = 0:`

`v` `= u + at`
`v_(t=3)` `= 0 + frac{100 sqrt3-120}{50} xx 3`
  `= 3.1923 \ …`
  `= 3.19 \ text{ms}^-1 \ text{(to 2 d.p.)}`