smc-1175-40-Equilibrium

Mechanics, SPEC2 2020 VCAA 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

`(mg)/T`
`T/(mg)`
`T/F`
`F/(mg)`
`(mg)/F`

Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces vertically:)`

`mg = Tsin(alpha)`

`text(Resolving forces horizontally:)`

`F = Tcos(alpha)`

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

`=>E`

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

`1/2 sin(alpha)`
`2sin(alpha)`
`1/2text(cosec)(alpha)`
`1/2cos(alpha)`
`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`

Mechanics, SPEC1 2019 VCAA 9

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.

`qquad qquad`

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

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

Show Answers Only

`T = (mg)/(2sinalpha)`
`text(See Worked Solutions)`

Show Worked Solution

`2 xx Tsinalpha`	`= mg`
`:.T`	`= (mg)/(2sinalpha)`

`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, SPEC2 2012 VCAA 22 MC

A 12 kg mass is suspended in equilibrium from a horizontal ceiling by two identical light strings. Each string makes an angle of 60° with the ceiling, as shown.

The magnitude, in newtons, of the tension in each string is equal to

A. `6 g`

B. `12 g`

C. `24 g`

D. `4 sqrt 3 g`

E. `8 sqrt 3 g`

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`D`

Show Worked Solution

`T/(sin 30^@)`	`= (12 g)/(sin 120^@)`
`2T`	`= (24 g)/sqrt3`
`T`	`= (12 g)/sqrt3`
	`= 4g sqrt 3`

`=> D`

Mechanics, SPEC1 2011 VCAA 7

A flowerpot of mass `m` kg is held in equilibrium by two light ropes, both of which are connected to a ceiling. The first rope makes an angle of 30° to the vertical and has tension `T_1` newtons. The second makes an angle of 60° to the vertical and has tension `T_2` newtons.

Show that `T_2 = T_1/sqrt 3.` (1 mark)
The first rope is strong, but the second rope will break if the tension in it exceeds 98 newtons.

Find the maximum value of `m` for which the flowerpot will remain in equilibrium. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`m = 20\ text(kg)`

Show Worked Solution

a. `text(Forces in equilibrium:)`

`tan 30^@`	`=T_2/T_1`
`T_2`	`= T_1 xx tan30^@`
`T_2`	`= T_1/sqrt3\ \ text(.. as required)`

b.	`sin30^@`	`= (T_2)/(mg)`
	`T_2`	`= (mg)/2`
	`98`	`=(m_text(max) xx 9.8)/2`
	`m_text(max)`	`=(2 xx 98)/9.8`
		`=20\ text(kg)`

Mechanics, SPEC2 2013 VCAA 16 MC

Forces of magnitude 5 N, 7 N and `Q` N act on a particle that is in equilibrium, as shown in the diagram below.

The magnitude of `Q`, in newtons, can be found by evaluating

A. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@))`

B. `5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@)`

C. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@))`

D. `5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@)`

E. `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(20^@))`

Show Answers Only

`A`

Show Worked Solution

`text(Using cosine rule:)`

`Q^2`	`= 5^2 + 7^2 – 2(5)(7)cos(70^@)`
`Q`	`= sqrt(5^2 + 7^2 – (5)(7)cos(70^@))`

`=> A`

Mechanics, SPEC2 2015 VCAA 16 MC

The diagram above shows a mass suspended in equilibrium by two light strings that make angles of `60^@` and `30^@` with a ceiling. The tensions in the strings are `T_1` and `T_2`, and the weight force acting on the mass is `underset~W`. The correct statement relating the given forces is

A. `underset~T_1 + underset~T_2 + underset~W = underset~0`

B. `underset~T_1 + underset~T_2 - underset~W = underset~0`

C. `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~0`

D. `underset~T_1 xx sqrt3/2 + underset~T_2 xx 1/2 = underset~W`

E. `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~W`

Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince equilibrium exists:)`

♦♦♦ Mean mark 26%.

`sumunderset~F = underset~(T_1) + underset~(T_2) + underset~W = underset~0`

`=> A`

Mechanics, SPEC2 2016 VCAA 14 MC

Two light strings of length 4 m and 3 m connect a mass to a horizontal bar, as shown below

The strings are attached to the horizontal bar 5 m apart.

Given the tension in the longer string is `T_1` and the tension in the shorter string is `T_2`, the ratio of the tensions `T_1/T_2` is

A. `3/5`

B. `3/4`

C. `4/5`

D. `5/4`

E. `4/3`

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`B`

Show Worked Solution

`text{Strings make up 3-4-5 (right-angled) triangle:}`

♦ Mean mark 41%.

`text(Let)\ \ theta =\ text(angle between 3m string and horizontal bar)`

`T_1/T_2`	`= cot theta`
	`= 3/4`

`=> B`

Mechanics, SPEC1 2016 VCAA 1

A taut rope of length `1 2/3` m suspends a mass of 20 kg from a fixed point `O`. A horizontal force of `P` newtons displaces the mass by 1 m horizontally so that the taut rope is then at an angle of `theta` to the vertical.

Show all the forces acting on the mass on the diagram below. (1 mark)

Show that `sin (theta) = 3/5`. (1 mark)
Find the magnitude of the tension force in the rope in newtons. (2 marks)

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`text(See Worked Solutions)`
`text(Proof)\ text{(See Worked Solutions)}`
`T = 245\ text(N)`

Show Worked Solution

b.	`sin(theta)`	`= 1/(1 2/3)`
		`= 1/((5/3)`
		`= 3/5`

`cos theta = (20g)/T`

`sin theta = 3/5\ \ text{(using part b)}`

`underbrace{(400g^2)/T^2 + 9/25}_(cos^2 theta + sin^2 theta = 1)`	`= 1`
`(400g^2)/T^2`	`= 16/25`
`(20g)/T`	`=4/5`
`T/(20g)`	`= 5/4`
`T`	`= 25g`
	`=245\ text(N)`

Mechanics, SPEC1 2014 VCAA 8

A body of mass 5 kg is held in equilibrium by two light inextensible strings. One string is attached to a ceiling at `A` and the other to a wall at `B`. The string attached to the ceiling is at an angle `theta` to the vertical and has tension `T_1` newtons, and the other string is horizontal and has tension `T_2` newtons. Both strings are made of the same material.

i. Resolve the forces on the body vertically and horizontally, and express `T_1` in terms of `theta`. (2 marks)
ii. Express `T_2` in terms of `theta`. (1 mark)
Show that `tan (theta) < sec (theta)` for `0 < theta < pi/2`. (1 mark)
The type of string used will break if it is subjected to a tension of more than 98 N.

Find the maximum allowable value of `theta` so that neither string will break. (3 marks)

Show Answers Only

i. `T_1 = 5g sec theta`
ii. `T_2 = 5g tan theta`
`text(Proof)\ text{(See Worked Solutions)}`
`theta_max = pi/3`

Show Worked Solution

a.i.

`text(Resolve vertically): \ 5g = T_1 cos theta`

`text(Resolve horizontally): \ T_1 sin theta = T_2`

`:. T_1 = 5g sec theta`

a.ii.	`T_2`	`= 5g sec theta sin theta`
		`= 5g tan theta`

b.	`cos theta in (0, 1), \ theta in (0, pi/2)`
	`sin theta in (0, 1), \ theta in (0, pi/2)`

`:. sin theta < 1, \ theta in (0, pi/2)`

♦ Mean mark part (b) 40%.

`text(If)\ sin theta < 1 and cos theta > 0:`

`(sin theta)/(cos theta) < 1/(cos theta)`

`:. tan theta < sec theta, quad 0 < theta < pi/2`

c. `tan theta < sec theta \ \ =>\ \ T_2\ text(will be smaller) => \ T_1\ text(will break first)`

♦ Mean mark part (c) 46%.

`5text(g)\ sec theta_max`	`= 98`
`sec theta_max`	`= 98/(5 xx 9.8)`
`sec theta_max`	`= 10/5`
`sec theta_max`	`= 2`
`:. theta_max`	`= pi/3`

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`

Mechanics, SPEC2 2018 VCAA 16 MC

The diagram below shows a mass being acted on by a number of forces whose magnitudes are labelled. All forces are measured in newtons and the system is in equilibrium.

The value of `F_2` is

A. `sqrt 2/2 (8 + 3 sqrt 3)`

B. `(11 sqrt 2)/2`

C. `(3 sqrt 2)/2`

D. `7.78`

E. `7.0`

Show Answers Only

`B`

Show Worked Solution

`text(vertical:)\ \ sum F_y = 4 + 3 sin 30^@ +\ ^(−)F_2 sin 45^@ = 0`

`4 + 3/2 – F_2/sqrt 2`	`= 0`
`F_2/sqrt 2`	`= 4 + 3/2`
`:. F_2`	`= sqrt 2 (4 + 3/2)`
	`= (11 sqrt 2)/2`

`=> B`