Question

one end of a uniform beam of mass 5 kg is mounted at the wall by hinges and the other end is held by a cable which is connected to the ceiling. The cable forms a 60 degree angle with the horizontal beam. Applying the equilibrium condition, find the force of tension at the cable and the vertical and horizontal components of the force of the hinge Fv and Fh on the beam and indicate their directions.

Answer 1

Since the system is in equilibrium thus Net torque of the system(due to the Tension in the cable and Weight of the beam) with respect to hinge will be 0.

Also Since the beam is uniform thus we can assume that the weight of the beam is acting at the centre of the beam. Let the length of the beam be "L" and "T" be the tension in the cable.

Then Net Torque=0

or

   (where W= Weight of the beam)

  

  

   ( where m= mass of beam and g= gravity)

  (ANS)

b). Now Using balance of Forces,

Horizontal Component of Force on hinge= Horizontal component of Tension in the cable

thus    (in outward direction following balance of forces)

(ANS)

And using balance of forces in vertical direction with F_v as the vertical component of force on hinge

  

(Considering downward direction as negative and upward direction as positive)

then

  

  (ANS)

(Negative sign shows that this vertical component of force is in downward direction)