Added new codeHEADmaster

1 files changed, 47 insertions, 0 deletions

diff --git a/3886/CH2/EX2.14/2_14.txt b/3886/CH2/EX2.14/2_14.txt
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+ --> // Finding magnitude of F

+--> //When F is applied at point B,refer fig 2.18(a)and(b)

+--> //From triangle AOC

+--> OC=300-150

+ OC  = 

+

+   150.

+--> AO=300

+ AO  = 

+

+   300.

+--> alpha=acosd(OC/AO)  //degree

+ alpha  = 

+

+   60.

+--> //from triangle AOB using geometry we get angle OBA=30 degree

+--> //Resolving the forces we get

+--> R=2000/cosd(30)  //N

+ R  = 

+

+   2309.4011

+--> F=R*sind(30)  //N

+ F  = 

+

+   1154.7005

+--> printf("Least force through point B is F=%.1f N",F)

+Least force through point B is F=1154.7 N--> //Least force required through the centre of roller

+--> //Assume that F makes an angle theta with the horizontal

+--> //Refer fig. 2.19 (a) and (b)

+--> //Resolving the forces we get

+--> //F*cosd(theta)=R*sind(60)...(1)

+--> //F*sind(theta)+R*cosd(60)=W...(2)

+--> //Solving (1) and (2) we get

+--> //sind(theta)+cotd(60)*cosd(theta)=W/F

+--> //For obtaining maximum value of W/F we differentiate W/F w.r.t. theta and we get

+--> theta=acotd(cotd(60))  //degree

+ theta  = 

+

+   60.

+--> //Least value of F is observed when it is at right angle to reaction R

+--> Fmin=2000*sind(60)  //N

+ Fmin  = 

+

+   1732.0508

+--> printf("\nLeast force through the centre of roller is Fmin=%.0f N",Fmin)

+

+Least force through the centre of roller is Fmin=1732 N
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