A non-uniform beam of length 4.41m and weight 1.02kg makes an angle of 25.2 with

Question

A non-uniform beam of length 4.41m and weight 1.02kg makes an angle of 25.2 with the horizontal. It is held in position by a frictionless pivot at its upper-right end and by a cable a distance of 3.01m farther down the beam and perpendicular to it. The center of gravity of the beam is a distance of 2.06m down the beam from the pivot. Lighting equipment exerts a downward force of 5.10kN on the lower-left end of the beam.

A) Find the tension T in the cable.

B)Find the vertical component of the force exerted on the beam by the pivot.

C) Find the magnitude of the horizontal component of the force exerted on the beam by the pivot.

Show work please so i can follow. Thanks a bunch!

Explanation / Answer

Weight of pole=mg=5100 N length of pole = L =4.41 m A vertical cable is attached a distance 3.01 m below its upper end The vertical cable is attached a distance 2.06 m above the lower end Assuming friction is ignored Suppose tension in the vertical cable = T Suppose force exerted by ground = F As the pole is in rotational and translational equilibrium, resultant force and resultant torque are zero As weight and tension are in vertically opposite direction and sum of the three forces F,T ,and mg is zero force F exerted by ground on the pole is in vertical upward direction suppose pole makes angle O with the vertical, perpedicular distance of line of action of tension tension (T) from end of pole on ground=x1=2.06 sinO weght acts at mid point of pole, mid point is 4.60 m from lower end perpedicular distance of line of action of weight (mg) from end of pole on ground=x2=3.01 sinO torque due to tension=T*2.06 sinO clockwise torque due to weight=mg*3.01 sinO anticlockwise As resultant torque is zero,clockwise torque=anticlockwise torque T*2.06sinO =5100*2.06 sinO T=5100*2.06/3.01 T =3490.36 N (A) tension in the cable is 3490.36 N B) F+T=mg F=mg -T= 5100 -3490.36=1609.63 N F=1609.63 N the magnitude of the force exerted by the ground on the pole. is1609.63 N C) As weight and tension are in vertically opposite direction and sum of the three forces F,T ,and mg is zero Since weight and tension are vertical, the third force F has to be vertical to give zero resultant force F exerted by ground on the pole is in vertical upward direction

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