Two rods are connected by a slider block D and are held in e

Two rods are connected by a slider block D and are held in equilibrium by the couple M_A as shown. Knowing that the coefficient of static friction between rod AC and the slider block is 0.40, write a computer program and use it to determine, for values of theta from 0 to 120 degree and using 10 degree increments, the range of values of MA for which equilibrium is maintained.

Solution

solution:

1)here closed loop equation is

AD\'+DB\'+BA\'=0

where in vector form it is given by

r1*e^im1+r2*e^im2+r3*e^im3=0

where angle m2=theta, goes on varying where m3=180 and m1 is given cosine rule as follows

where length of link1 AD is

r1=(r2^2+r3^2-2*r3*r2*cosm2)^.5

which gives r1=100 mm

where angle between perpendicular to link 1 and link 2 is given by

m4=180-m2-m1

where for real part of closed loop equation is

r1*cosm1+250cosm2+150cos180=0

m1=arccos[150-250cosm2/r1]

where torque in second link gives tangential force and part of force perpendicular to link 1 is responsible is given by

M2=F2*r2

F1=2.5*1000/250=10 N

where force perpendicular to link 1 is

Ft=F1cosm4

and moment in first link 1 is

M1=(Ft)*r1

for all value of M1 there is equillibrium

hence matlab code is as follows

clc;

clear all;

close all;

m3=pi;

F1=10;

r2=250;

r3=150;

for m2=0:0.1745:2.094

r1=sqrt(r2^2+r3^2-2*r3*r2*cos(m2));

m1=arccos[(150-250cos(m2))/r1];

m11=m1*pi/180;

m4=pi-m2-m11;

Ft=F1*cos(m4);

M1=Ft*r1;

fprintf(\'\ moment is =%f\',M1);

end