weight of beam m1*g*L/2(centre of mass at L/2 assuming uniform distrubution of mass)

PPhheewww

Here i go.

The beam is hanging from the building, hinged to it. The positioning of the beam is horizontal as it is supported by a string at the other end and pulled up.

Let the tension be T

Mass of beam m1

Mass of man m2

Force exerted be hinge fx and fy.

Now, let your DATUM or point of reference be the hinge.

For the beam to be static along the vertical direction, the net force and torque along vertical has to be zero.

TORQUE DUE TO EQUAL TO

So, the upward torque should be equal to downward torque.

Hence

fy(hinge along y) fy*0(reference is hinge)

PPhheewww

Here i go.

The beam is hanging from the building, hinged to it. The positioning of the beam is horizontal as it is supported by a string at the other end and pulled up.

Let the tension be T

Mass of beam m1

Mass of man m2

Force exerted be hinge fx and fy.

Now, let your DATUM or point of reference be the hinge.

For the beam to be static along the vertical direction, the net force and torque along vertical has to be zero.

m2*g*d m1*g*L/2=Tsin(theta)*L

So, the upward torque should be equal to downward torque.

Hence

So, the upward torque should be equal to downward torque.

Hence

For the horizontal,

Lcos(theta)=fx for it to not to move along x axis.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022