Question

A mechanic can open a nut by applying a force of 120N while using a lever handle of length 40cm. What is the length of the handle required if he wants to apply the force of only 30N.​

Answer 1

$\huge \underline{ \underline{ { \cal \: {★Q}} \large \red{ \tt \: U} \green{ \tt \: E} \blue{ \tt \: S} \orange{ \tt \: T} \purple{ \tt \: I} \red{ \tt \: O} \blue{ \tt \: N}}} : -$

A mechanic can open a nut by applying a force of 120N while using a lever handle of length 40cm. What is the length of the handle required if he wants to apply the force of only 30N.

$\underline{ \huge{ \sf \: \ ☯\:\: T} \large\tt{{{ \underline{ \blue{HINGS}}}}}}$$\underline{ \huge{ \sf \: \ G}\large\tt{{{ \underline{ \blue{IVEN} : - }}}}}$

$\boldsymbol { \underline{★Force \: \: required \: \: to \: \: open \: \: a \: \: nut \: \: \orange➙ \: \red{\tt {\: 120 \: N}}}}$

$\boldsymbol { \underline{★Length \: \: of \: \: the \: \: levers \: \: handle \: \: \orange➙ \: \red{\tt {\: 40 \: \bf{cm }}}}}$

$\boldsymbol { \underline{★Force \: \: required \: \: to \: \: open \: \: the \: \: nut \: \: of \: \: another \: \: handle \: \orange➙ \: \red{\tt {\: 30 \: N}}}}$

$\underline{ \huge{ \sf \: \ ☯\:\: T} \large\tt{{{ \underline{ \pink{O}}}}}}$$\underline{ \huge{ \sf \: \ F}\large\tt{{{ \underline{ \pink{IND} : - }}}}}$

$\boldsymbol { \underline{★Length \: \: of \: \: the \: \: lever \: \: of \: \:another\: \: handle \: \: \orange➙ \: \red{\tt {\:??}}}}$

$\underline{ \huge{ \sf \: \ ☯ \:\:F} \large\tt{{{ \underline{ \purple{ORMULA}}}}}}$$\underline{ \huge{ \sf \: \ U}\large\tt{{{ \underline{ \purple{SED} : - }}}}}$

$\boxed{⚫{\bf{\red{\underline{\: \:Note\: :- \: (i) \: means \: case \: 1, \: (ii) \:means \:case\: 2\:. }}}}}$

$\sf{★Effort\: arm\: (ii) =\dfrac {Effort\: applied\: (i) \times \:Handle's\: length\: (i)}{ Effort\: applied\: (ii)}}$

$\huge \underline{ \underline{ { \cal \: {★S}} \large \red{ \tt \: O} \green{ \tt \: L} \blue{ \tt \: U} \orange{ \tt \: T} \purple{ \tt \: I} \red{ \tt \: O} \blue{ \tt \: N}}} : -$

$\large\frak{\underline{\dag\:The \:work\: done \:to \:open \:the\:nut \:in \:both \:the\: cases\: is\:same.}}$

$\large\sf{\underline{\therefore\: \:work\: done \:in \:case \:(i):\implies \:work\: done \:in \:case \:(ii)}}$

$\huge { \underline{ { \sf \: {•\:\;T}} \large \blue{ \bf{ HEN}, }}}$

$\large\sf{\underline{★Effort\: \times \:effort\: arm\: in\:(i)\: \implies \:effort \: \times \:effort\: arm\: in \:case\: (ii)}}$

$\sf{\underline{★120\:N \times \:40\: cm\: in\:(i)\: \implies \:30 \:N \times \:Effort\: arm\: in \:case\: (ii)}}$

$\sf{\therefore\:\:Effort\: arm\: (ii) =\dfrac {120 \: \times\: 40\: }{ 30}}$

$\frak{\underline{\dag\:Multiply \:120\: and\: 40 \:to \:get \:4800}}$

$\sf:{\implies{\dfrac{4800}{30}}}$

$\frak{\underline{\dag\:Expand\: \dfrac{4800}{30}}}$

$\sf:{\implies{\dfrac{\cancel{4800}\:^{160}}{\cancel{30}\:^{1}}}}$

$\sf:{\implies{\dfrac{4800}{30} \approx 160}}$

$\huge \underline{ \underline{ { \cal \: {★A}} \large \red{ \tt \: N} \green{ \tt \: S} \blue{ \tt \: W} \orange{ \tt \: E} \purple{ \tt \: R} }} : -$

$\boxed{\sf{★\:\:\: \bigg\lgroup { \red{\dfrac{4800}{30} \approx 160\:\: } \bigg\rgroup★}}}$

$\huge \underline{ \underline{ { \cal \: {♪}} \large \red{ \tt \: ♪} \green{ \tt \: •} \blue{ \tt \: •} \orange{ \tt \: •} \purple{ \tt \: ✓} \red{ \tt \: ✓} \blue{ \tt \: ✓}}}$

Answer 2

HEYYA MATE

HERE'S YOUR ANSWER

ANSWER:-

~•in the first case force applied is

F= 120 N..

and magnitude of perpendicular distance (called the radius vector)

r=40cm...... =40/100m

Torque=120×40/100

~•in the second case force applied is

F=40N

and the length of the radius vector is to be found

r=?

Torque=F×r is the same in the both cases

48=40×r

r=1.2m hence to apply a force of 40 N for the same torque, The mechanic should use a spanner of length 1.2 m.

Hope it helps plz tag it a brainliest answer