Balancing the Forces on a Steel Beam Construction

AI Thread Summary
The discussion centers on calculating the torque on a 4m steel beam with a 500 kg weight, bolted at one end, while a 70 kg construction worker stands at the opposite end. Participants clarify that two torques are acting: one from the worker's weight at the beam's end and another from the beam's weight at its center of gravity. To find the torque, the weights of both the beam and the worker must be calculated using their masses. The perpendicular distance for the worker's weight is the full length of the beam, while the distance for the beam's weight torque is half the beam's length. Understanding these distances and forces is crucial for accurately determining the total torque about the pivot point.
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Homework Statement


A 4m long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70kg construction worker stands at the far end of the beam


Homework Equations


What is the magnitude of the torque about the point where the beam is bolted into place?


The Attempt at a Solution


i thought about using (Absolute torque)=d1F+d2F=(d1+d2)F=lF (l=the distance between the lines of action person to beam)

but i have doubts that is the right one, any help is appreciated
 
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dals2002 said:

Homework Statement


A 4m long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70kg construction worker stands at the far end of the beam


Homework Equations


What is the magnitude of the torque about the point where the beam is bolted into place?


The Attempt at a Solution


i thought about using (Absolute torque)=d1F+d2F=(d1+d2)F=lF (l=the distance between the lines of action person to beam)

but i have doubts that is the right one, any help is appreciated
Ther are two torques acting, one by the person's weight acting at the end of the beam, and the other from the beam's weight acting at its cg.
 
yea that's the part i didn't get, i mean i know there is 2 torques that's why i thought of the equation above, but besides that i didn't understand anything else

t=rFt but i don't have the force either and they only gave me mass and distance, so I am unsure on what step to take next
 
dals2002 said:
yea that's the part i didn't get, i mean i know there is 2 torques that's why i thought of the equation above, but besides that i didn't understand anything else

t=rFt but i don't have the force either and they only gave me mass and distance, so I am unsure on what step to take next
You have the mass of the beam and the mass of the person. There is a simple formula that converts mass to weight. Can you use it to find the weights? These weights are the forces you are looking for. Then you need to know the perpendicular distances of the line of action of each of these forces to the pivot point at the bolted end of the beam, in order to calculate the torques.
 
ok i understand all of it, but i have one last question how can i find the perpendicular distance of the line to the pivot?

Thanks for all your help
 
dals2002 said:
ok i understand all of it, but i have one last question how can i find the perpendicular distance of the line to the pivot?

Thanks for all your help
The persons weight acts vertically down at the end of the beam. The beam is horizontal. So the perpendicular distance from the vertical weight force line of action to the pivot is just the beam length of 4m. Now, what distance should you use in calculating the beam weight torque about the pivot?
 
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