the force applied at the handle of the strict lever causes thelever to revolve clockwise about the pen B v an angle of 2degrees. Determine the mean normal strain emerged in eachwire. The wires space unstretched once the bar is in the horizontalposition. Please display work.

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the force applied at the manage of the rigid lever reasons the bar to rotate clockwise about the pin B v an edge of 2 degrees. Determine the typical normal strain arisen in every wire. The wires space unstretched when the lever is in the horizontal position. Please display work.

the force applied at the manage of the rigid lever reasons the bar to rotate clockwise about the pin B v an edge of 2 degrees. Determine the typical normal strain arisen in every wire. The wires space unstretched when the lever is in the horizontal position. Please display work.

**Angular displacement**: that the as the angle through which an object moves top top a one path. It is the angle, in radians, between the initial and also final positions.

**Strain:** The change in measurement over the original measurement caused due to the used load is dubbed strain.

Strain is the ratio of adjust in length over the initial length.

ε=δL\\varepsilon = \\frac\\delta Lε=Lδ

The relation for angular displacement is,

δ=Lθ\\delta = L\\theta δ=Lθ

Conversion of level to radians,

θ=2∘(π180)rad=0.03491rad\\beginarrayl\\\\\\theta = 2^\\circ \\left( \\frac\\pi 180 \\right)\\;\\rmrad\\\\\\\\\\;\\;\\;\\rm = 0\\rm.03491\\;\\rmrad\\\\\\endarrayθ=2∘(180π)rad=0.03491rad

Calculate the displacement of point A.

δA=LABθ\\delta _A = L_AB\\theta δA=LABθ

Here size of lever abdominal is LABL_ABLAB and angle that tilt about the pen B is θ\\theta θ .

Substitute, 200mm200\\;\\rmmm200mm because that LABL_ABLAB , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,

δA=(200×0.03491)=6.9813mm\\beginarrayl\\\\\\delta _A = \\left( 200 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 6.9813\\;\\rmmm\\\\\\endarrayδA=(200×0.03491)=6.9813mm

Calculate the displacement of suggest C.

δC=LBCθ\\delta _C = L_BC\\theta δC=LBCθ

Here size of bar BC is LBCL_BCLBC and also angle of tilt around the pen B is θ\\theta θ .

Substitute, 300mm300\\;\\rmmm300mm for LBCL_BCLBC , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,

δC=(300×0.03491)=10.4720mm\\beginarrayl\\\\\\delta _C = \\left( 300 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 10.4720\\;\\rmmm\\\\\\endarrayδC=(300×0.03491)=10.4720mm

Calculate the displacement of suggest D.

δD=LBDθ\\delta _D = L_BD\\theta δD=LBDθ

Here length of lever BD is LBDL_BDLBD and also angle the tilt around the pin B is θ\\theta θ

Substitute 300mm300\\;\\rmmm300mm for LBDL_BDLBD , 0.03491rad\\rm0\\rm.03491\\;\\rmrad0.03491rad for θ\\theta θ in relation,

δD=(500×0.03491)=17.4533mm\\beginarrayl\\\\\\delta _D = \\left( 500 \\times 0.03491 \\right)\\\\\\\\\\;\\;\\;\\; = 17.4533\\;\\rmmm\\\\\\endarrayδD=(500×0.03491)=17.4533mm

Calculate the typical normal strain occurred in cable AH.

(∈avg)AH=δALAH\\left( \\in _avg \\right)_AH = \\frac\\delta _AL_AH(∈avg)AH=LAHδA

Here displacement of allude A of the lever is δA\\delta _AδA , unstretched length of the cable AH is LAHL_AHLAH .

Substitute 6.9813mm6.9813\\;\\rmmm6.9813mm because that δA\\delta _AδA , 200mm200\\rm mm200mm because that LAHL_AHLAH in relation,

(∈avg)AH=6.9813200=0.0349\\beginarrayl\\\\\\left( \\in _avg \\right)_AH\\; = \\frac6.9813200\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0349\\\\\\endarray(∈avg)AH=2006.9813=0.0349

Calculate the mean normal strain developed in wire CG.

(∈avg)CG=δCLCG\\left( \\in _avg \\right)_CG = \\frac\\delta _CL_CG(∈avg)CG=LCGδC

Here displacement of suggest C that the bar is δC\\delta _CδC , unstretched length of the wire CG is LCGL_CGLCG

Substitute 10.4720mm10.4720\\;\\rmmm10.4720mm for δC\\delta _CδC and 300mm300\\rm mm300mm because that LCGL_CGLCG in relation,

(∈avg)CG=10.4720300=0.0349\\beginarrayl\\\\\\left( \\in _avg \\right)_CG\\; = \\frac10.4720300\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0349\\\\\\endarray(∈avg)CG=30010.4720=0.0349

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**Calculate the median normal strain emerged in wire DF.See more: Viral Video Of 'The Star Spangled Banner As You Ve Never Heard It**

** (∈avg)DF=δDLDF\\left( \\in _avg \\right)_DF = \\frac\\delta _DL_DF(∈avg)DF=LDFδD **

**Here displacement of suggest D of the lever is δD\\delta _DδD , unstretched size of the wire DF is LDFL_DFLDF .**

**Substitute, 17.4533mm17.4533\\;\\rmmm17.4533mm because that δD\\delta _DδD and also 300mm300\\rm mm300mm for LDFL_DFLDF in relation,**

** (∈avg)DF=17.4533300=0.0582\\beginarrayl\\\\\\left( \\in _avg \\right)_DF = \\frac17.4533300\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = 0.0582\\\\\\endarray(∈avg)DF=30017.4533=0.0582 **

**Ans:**