Question

$\huge{\boxed{\bf{Hola \ Amigos!}}}$
Physics Question, chapter Dynamics

A metre scale 'PQ' of uniform thickness and weighing 100 $\rm{g_{wt}}$ is suspended by means of a string to a rigid support at a point 'R' on it such that $\boxed{\sf{PR: QR = 2:3}}$ . At the end 'P' a load of 85 $\sf{g_{wt}}$ is attached.

1. Find the load attached at Q such that the scale is in equilibrium.

2. If 'S' is a point on the scale such that $\boxed{\sf{PR: SR = 2:1}}$ and the load at 'Q' is shifted to 'S', find the percentage change in and the the load, in shifting it from 'Q' and 'S'; to maintain equilibrium state of the scale.

Kinda muddled, please help :D​

Answer 1

$\huge\bf{\underline{\color{indigo}GIVEN}}$

☞︎︎︎ First of all, here we elaborate the diagram of this problem i.e. attached in part-1 (See the attachment diagram.) .

☯︎ It is given that a metre scale PQ of uniform thickness and weighing $\bf{100~g_{wt}}$ is suspended by a string. And the point at which the scale is attached that point be ‘R’. At point ‘P’ a load of $\bf{85~g_{wt}}$ is attached.

And consider that at point ‘Q’ a load of $\bf{m_o~g_{wt}}$ is attached .

Now it is given that,

$\pink\longmapsto~\bf\gray{PR~:~QR~=~2~:~3~}$

So,

Consider that $\bf{\red{PR~=~2L~\&~QR~=~3L}}.$

$\huge\bf{\underline{\color{orange} SOLUTION}}$

$:\implies~\bf{\dfrac{PR}{QR}~=~\dfrac{2}{3}}$

$:\implies~\bf{PR~=~\dfrac{2}{3}~QR}$

$:\implies~\bf{QR~=~\dfrac{3}{2}~PR}--(i)$

☯︎ PQ is a metre scale. So the length of the scale is 100cm.

➝ PQ = 100 cm

➝ PR + QR = 100 cm

➝ PR = 100 - QR

➝ PR = 100 - $\rm{\dfrac{3}{2}\:PR~}$ (from i)

➝ PR + $\rm{\dfrac{3}{2}\:PR~}$ = 100

➝ $\rm{\dfrac{5}{2}\:PR~}$ = 100

➝ PR = $\sf{100\times{\dfrac{2}{5}}~}$

➝ PR = 40 cm

➝ QR = 60 cm

☯︎ If we put “PQ = $\bf{100~g_{wt}}$”, then

➝ PR = $\bf{40~g_{wt}}$

➝ QR = $\bf{60~g_{wt}}$

✈︎ Now, for the scale to be equilibrium

$:\implies~\bf{\Big(40\times{\dfrac{2L}{2}}\Big)~+~85\times{2L}~=~\Big(60\times{\dfrac{3L}{2}}\Big)~+~m_o\times{3L}~}$

$:\implies~\bf{40L~+~170L~=~30\times{3L}~+~3m_o~L~}$

$:\implies~\bf{210L~=~90L~+~3m_o~L~}$

$:\implies~\bf{3m_o~L~=~210L~-~90L}$

$:\implies~\bf{3m_o~L~=~120L~}$

$:\implies~\bf{m_o~=~\dfrac{120L}{3L}~}$

$:\implies~\bf\green{m_o~=~40~g_{wt}~}$

$\huge\red\therefore$ The load attached at Q is weighing $\bf{\pink{40~g_{wt}}}.$

☯︎ Now, it is given that S is a point on the scale such that $\bf\gray{PR~:~SR~=~2~:~1~}$ and the load at Q is shifted to S. (See the attachment diagram part-2.)

So,

Consider that $\bf{\red{PR~=~2L~\&~SR~=~L}}.$

$:\implies~\bf{\dfrac{PR}{SR}~=~\dfrac{2}{1}}$

$:\implies~\bf{SR~=~\dfrac{1}{2}\times{PR}}$

➪ If PR = 40 cm,

$:\implies~\bf{SR~=~\dfrac{1}{2}\times{40}}$

$:\implies~\bf{SR~=~20cm~}$

➪ If PR = $\bf{40~g_{wt}}$

$:\implies~\bf{SR~=~20~g_{wt}~}$

☯︎ So it is sure that the load at Q is shifted towards the point R.

✈︎ If $\bf{m'_o}$ is the mass of load attached at S, then for the scale to be equilibrium

$:\implies~\bf{85\times{2L}~-~\Big(40\times{\dfrac{2L}{2}}\Big)~=~m'_o\times{L}~+~\Big(20\times{\dfrac{L}{2}}\Big)~}$

$:\implies~\bf{170L~-~40L~=~m'_o\times{L}~+~10L~}$

$:\implies~\bf{130L~=~m'_o\times{L}~+~10L~}$

$:\implies~\bf{130L~-~10L~=~m'_o\times{L}~}$

$:\implies~\bf{120L~=~m'_o\times{L}~}$

$:\implies~\bf\green{m'_o~=~120~g_{wt}~}$

Now,

$\longmapsto~\bf{Change~in~\%~=~\dfrac{120~-~40}{40}\times{100}~}$

$\longmapsto~\bf{Change~in~\%~=~\dfrac{80}{40}\times{100}~}$

$\longmapsto~\bf{Change~in~\%~=~2\times{100}~}$

$\longmapsto~\bf\purple{Change~in~\%~=~200\%~}$

$\huge\red\therefore$ The load at point S, which is shifted from Q is weighing $\bf{\pink{120~g_{wt}}}.$ And the change in percentage is 200% .

Answer 2

$\huge{\underline{\sf Given:-}}$

A metre scale 'PQ' of uniform thickness and weighing 100$\sf g_{wt}$ is suspended by means of a string to a rigid support at a point 'R' on it such that PR:QR = 2:3. At the end 'P' a load of 85$\sf g_{wt}$ is attached.

$\huge{\underline{\sf Find:-}}$

❏Load attached at Q such that the scale is in equilibrium.

❏The percentage change in and the the load, in shifting it from 'Q' and 'S'; to maintain equilibrium state of the scale.

$\huge{\underline{\sf Diagram:-}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier( - 0.9,0)(0,0)(5,0) \qbezier(1,0)(1,1)(1,1)\qbezier(0.5,1)(0,1)(1.5,1) \put( - 0.9,0){\vector(0,-1){1.5}}\put(2,0){\vector(0,-1){1.5}}\put(5,0){\vector(0,-1){1.5}}\multiput(3,0)(0,-0.3){5}{\line(0,-1){0.1}} \put(3,-1.4){\vector(0,-1){0}} \put(-1,-1.8){\bf 85} \put(1.7,-1.8){\bf 100} \put(2.8,-1.8){ \bf X^1 }\put(4.9,-1.8){\bf X} \put(-1,0.1){\bf P} \put(1.1,0.1){\bf R}\put(1.9,0.1){\bf T} \put(2.9,0.1){\bf S} \put(4.8,0.1){\bf Q} \end{picture}$

$\huge{\underline{\sf Solution:-}}$

Here, T is a midpoint of the scale and it's weight is 100g acting at this point.

Given,▶PR:QR = 2:3

$\sf \mapsto \dfrac{PR}{QR} = \dfrac{2}{3}$

$\sf \mapsto QR= \dfrac{3}{2} PR \qquad \bigg\lgroup Equation \: 1 \bigg\rgroup$

and, PR + QR = 100cm

$\sf \mapsto PR + QR =100$

$\sf \mapsto PR =100 - QR$

$\sf \mapsto PR =100 - \dfrac{3}{2} PR \qquad \bigg\lgroup using \: equation \: 1 \bigg\rgroup$

$\sf \mapsto PR + \dfrac{3}{2} PR =100$

☯$\underline{Taking\:L.C.M}$

$\sf \mapsto \dfrac{2PR + 3 PR}{2}=100$

$\sf \mapsto \dfrac{5PR}{2}=100$

$\sf \mapsto PR=100 \times \dfrac{2}{5}$

$\sf \mapsto PR=\dfrac{200}{5}$

$\sf \mapsto PR=40cm \qquad \bigg\lgroup Equation \: 2 \bigg\rgroup$

$\qquad\quad$ ____________________

$\sf \leadsto PR + QR =100$

$\sf \leadsto 40 + QR =100 \qquad \bigg\lgroup using \: quation \: 2\bigg\rgroup$

$\sf \leadsto QR =100 - 40$

$\sf \therefore QR =60cm$

If 'x' is the load attached at Q, then for the scale to be in equilibrium

➠(85g) (PR) = x(QR) + RT(100)

⇛(85)(40) = (x)(60) + 1000

⇛3400 -1000 = 60x

⇛2400/60 = x

⇛x = 40cm

$\small{\sf \therefore \underline{Load\: attached\:at\:Q \:is\:40cm}}$

════════════════════════════════════

Given, ▶PR:SR = 2:1

$\sf \mapsto \dfrac{PR}{SR} = \dfrac{2}{1}$

$\sf \mapsto SR= \dfrac{PR}{2}$

$\sf \mapsto SR= \dfrac{40}{2} \qquad \bigg\lgroup using \: quation \: 2\bigg\rgroup$

$\sf \mapsto SR=20cm$

Let, x¹ be the load attached at S.

Then (85)PR = (x¹)(SR) + 100(TR)

= (85)(40) = x'¹ (20) + (100)10

⇒x'¹ = 120g

If the load at Q is shifted to S, the percentage change in the load

$\dashrightarrow\sf \bigg\lgroup \dfrac{x^1 - x}{x} \bigg\rgroup 100$

$\dashrightarrow\sf \bigg\lgroup \dfrac{120-40}{40} \bigg\rgroup 100$

$\dashrightarrow\sf \bigg\lgroup \dfrac{80}{40} \bigg\rgroup 100$

$\dashrightarrow\sf \bigg\lgroup \dfrac{8}{4} \bigg\rgroup 100$

$\dashrightarrow\sf 2 \times 100$

$\dashrightarrow\sf 200\%$

$\small{\therefore\sf \underline{Percentage \: change \: in \: the \: load \: is \: 200\%}}$