A uniform beam of mass MB = 0.
A uniform beam of mass MB = 0.8833 kg and length l = 0.8m isattached to a wall by a hinge at at one end and by a cord ofnegligible mass and length ls = 1.27 m that makes an angle ? withthe horizontal at the other end (see figure drawn on the board),such that the beam is horizontal. A mass mH = 0.55 kg is hung fromthe beam at a distance lH = 0.6 m from the hinge. What is thetension in the cord, and what is the magnitude and direction of theforce between the hinge and beam?
Answer:
a) angle made by cord with the beam, theta =cos^-1(0.8/1.27)
= 50.96 degrees
As the beam is in equilibrium net force and net torqueacting on the beam must be zero.
let T is the tension in the cord.
net torque about hinge = 0
0.8333*9.8*(0.8/2) + 0.55*9.8*0.6*sin(90) -T*0.8*sin(50.96) = 0
T = (0.8333*9.8*(0.8/2) +0.55*9.8*0.6*sin(90))/(0.8*sin(50.96))
= 10.5 N<<<<<<<————-Answer
b) let Fx and Fy are component of force exerted by thehinge.
use, Fnetx = 0
Fx – T*cos(50.96) = 0
Fx = T*cos(50.96)
= 10.5*cos(50.96)
= 6.61 N
Apply, Fnety = 0
Fy + T*sin(50.96) – 0.8833*9.8 – 0.55*9.8 =0
Fy = 0.8833*9.8 + 0.55*9.8 -10.5*sin(50.96)
= 5.89 N
F_hinge = sqrt(Fx^2 +Fy^2)
= sqrt(6.61^2 + 5.89^2)
= 8.85 N<<<<<<<<<<———————-Answer
direction : theta = tan^-1(Fy/Fx)
= tan^-1(5.89/6.61)
= 41.7 degrees above +x axis<<<<<<<<<<———————-Answer
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