Question

A beam of length L is attached to a string as shown in picture. The mass of the beam is M=20kg. Determine the tension in the string and the force on the beam exerted by the pivot point.

Answer 1

let tension in the string be T, acting in horizontal direction to the right

let force on the beam exerted by the pivot be :

Py, in vertical direction , in upward direction

Px, in horizontal direction, to the left direction

weight of the beam is acting at midpoint of the beam with magnitude of 20*9.8=196 N, in vertically downward direction

hence balancing force along vertical direction:

Py=weight of the beam=196 N

balancing force in horizontal direction:

Px=T

now to find out value of T:

balancing net toruqe about the pivot point:

perpendicular distance of tension T from pivot=(L/3)*sin(30)=L/6

weight of the beam acts at a distance of L/2 from the end point of the beam

hence distance across the beam from the pivot=(L/2)-(L/3)=L/6

then perpendicular distance from the pivot =(L/6)*cos(30)=0.14434*L

hence balancing torque about pivot:

T*L/6=weight*0.14434*L

==>T=weight*0.14434*6=169.74 N

hence Px=169.74 N

so final solution :

tension=169.74 N

force from pivot:

horizontal force=169.74 N

vertical force=196 N

net force magnitude=259.28 N