Pravshin90

Hai Pravs, Welcome to wikipedia

Hello Pravallika krishna, Welcome to wikipedia. Hope that you contribute as well learn a lot from wikipedia. Enjoy wikiing

Ganeshsashank (talk) 06:48, 26 February 2010 (UTC)

Problem 4.3.5

Watch the figure. The question is

Determine the forces in each bar of the truss shown caused by lifting of 120lb load at constant velocity.

Ans) The pulley is being lifted with constant velocity. So no acceleration. Hence no net force.

Consider the FBD of pulleysystem at C. So as to lift the 120lb weight without acceleration,

$Pcos\theta =120\,\!$

But $cos\theta ={\frac {3}{5}}\,\!$

Hence $Psin\theta ={\frac {120\times 4}{3}}$

$=160\,\!$

Hence, C is acted upon by an horizontal force of 160lb.

Now consider $\sum F_{x}=0\,\!$ and $\sum F_{y}=0\,\!$ If $A_{v}$ and $D_{v}$ are the vertical reactions at A and D respectively and $D_{h}$ is the horizontal reaction at D, then,

$A_{v}+D_{v}=120+Pcos\theta \,\!$

$=240lb\,\!$

$D_{h}=Psin\theta \,\!$

$=160lb\,\!$

Clearly $A_{v}=D_{v}=120lb\,\!$

Now consider forces in bars AB,BC,CD,AC,BD be ${\overrightarrow {AB}},{\overrightarrow {BC}},{\overrightarrow {CD}},{\overrightarrow {AC}},{\overrightarrow {BD}}\,\!$

Consider FBD of A:

After resolving force, we get....

$A_{v}=BAsin\theta \,\!$

$AC=-BAcos\theta \,\!$

Consider FBD of D:

After resolving, we get....

$D_{v}=BDsin\theta \,\!$

$D_{h}=CD+BDcos\theta \,\!$

Consider FBD of B:

After Resolving, we get....

$BD=BA\,\!$

$BC=-2\times BDsin\theta \,\!$

Similarly Consider FBD of C: Again resolving we get...

$AC=-CD\,\!$

So, Finally we get

Member	Force
AB	200lb
AC	32 lb
BC	240lb
BD	200lb
CD	32 lb

Ganeshsashank (talk) 07:34, 26 February 2010 (UTC)

C program.

1. include<stdio.h>

main()

{

int m1,m2,m3,sum,p;

float avg,percent;

printf("Total marks per subject is 100\n");

printf("Enter three subject marks:\n");

scanf("%d%d%d",&m1,&m2,&m3);

sum=m1+m2+m3;

avg=sum/3;

percent=sum*100;

percent= percent/300;

p=percent/10;

printf("sum=%d\t Average=%f\t Percent=%f",sum,avg,percent);

switch(p)

{

case 7: printf("Person got A\n");

break;

case 8: printf("Person got A\n");

break;

case 9: printf("Person got A\n");

break;

case 6: printf("person got B\n");

break;

case 5: printf("person got C\n");

break;

case 4: printf("Person got D\n");

break;

default: printf("The person got E\n");

break;

}

}

Hey pravs, got an excellent question.Solve it.

File:Equilibrium of force-sample problem.jpg

Q). A uniform ring of mass m has a radius r carries an eccentric mass $'''m_{0}'''$ at radius b and is in equilibrium position on an incline, which makes an angle $'''\alpha '''$ with the horizontal. If the contacting surfaces are rough enuogh to prevent slipping, write the expression for the angle $'''\theta '''$ which defines the equilibrium position.

Ans)$\theta =sin^{-1}[{\frac {r}{b}}(1+{\frac {m}{m_{0}}})sin\alpha ]\,\!$ Ganeshsashank (talk) 07:00, 27 February 2010 (UTC)