Torque/Static Equilibrium/Forces problem

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AI Thread Summary
The discussion revolves around a static equilibrium problem involving a rigid beam supported at both ends, with a student standing on it. The key issue is determining the upward force exerted by support 1 while considering the torques acting on the beam. Participants clarify that the weight of the beam should be treated as acting at its center, and the student’s weight is the only additional force to consider for calculating torque at support 1. A free body diagram is recommended to visualize the forces, and it is confirmed that support 2 can be ignored for this calculation. The user ultimately resolves their confusion after receiving guidance on correctly applying the principles of torque and static equilibrium.
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Torque/Static Equilibrium/Forces problem...

I have been working on this problem for a few hours, but I just can't get it ...

Homework Statement

Problem:

A 3.0-m-long rigid beam with a mass of 100 kg is supported at each end. An 80 kg student stands 2.0 m from support 1. How much upward force does support 1 exert on the beam?

here is an image:

http://photo.ringo.com/208/208651783O865465193.jpg

----

Homework Equations



τ=Frsinφ
F=ma

The Attempt at a Solution



I figured out the torques kinda, but do i use the mass as (2/3*100)+80? or (2/3*100)+40...either way i kept getting a wrong answer...

and the center of mass i assume is where the person is right? bec the 80kg addition would be the most massive point right?

since the angles are 90 degrees, the sine of φ doesn't matter right?, i mean it would only affect the sign +/- of the answer...since τ=Frsinφ, and φ is either 90 or -90 degrees...

idk...basically i just can't get this...

i got solutions ranging from 76.666.., 1306.666... etc...
 
Last edited by a moderator:
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Assume the weight of the beam acting at the center of the beam, and set the sum of all torques with respect to support B equal zero.
 
radou said:
Assume the weight of the beam acting at the center of the beam, and set the sum of all torques with respect to support B equal zero.

so do i pretend support 2 doesn't exist? and then just find the torque for support 1 using [80kg+(2/3*100)] as the mass?
 
Last edited:
aliaze1 said:
so do i pretend support 2 doesn't exist? and then just find the torque for support 1 using [80kg+(2/3*100)] as the mass?

Yes, you can pretend support 2 doesn't exist. Do you know how to draw a free body diagram? You have three forces acting - one is the weight of the bar, the second one is the weight of the person, and the third one is the force at support 1 you need to find.
 
lol...yea i got it, thanks, i went to a tutor today bec i was making small errors...i for the torque of the person, i was using the weight as (person+segment of rod), when its only supposed to be person

thanks for the help, greatly appreciated
 

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