Question

A uniform beam of mass MB = 0.8833 kg and length l = 0.8m is attached to a wall by a hinge at at one end and by a cord of negligible mass and length ls = 1.27 m that makes an angle ? with the 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 from the beam at a distance lH = 0.6 m from the hinge. What is the tension in the cord, and what is the magnitude and direction of the force between the hinge and beam?

Answer 1

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 torque acting 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 the hinge.

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