# 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?

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))

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)

direction : theta = tan^-1(Fy/Fx)

= tan^-1(5.89/6.61)

= 41.7 degrees above +x axis<<<<<<<<<<———————-Answer

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