Torque/Static Equilibrium/Forces problem

  • Thread starter aliaze1
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In summary, the problem involves a 3.0-m-long beam with a mass of 100 kg and an 80 kg student standing 2.0 m from support 1. The goal is to find the upward force exerted by support 1 on the beam. The equations used are τ=Frsinφ and F=ma, and the weight of the beam is assumed to act at the center of the beam. The solution involves finding the torque of the person by using their weight as the mass and drawing a free body diagram to identify the three forces acting on the beam. After correcting small errors with the torque calculation, the correct answer was found with the help of a tutor.
  • #1
aliaze1
174
1
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...
 
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  • #2
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.
 
  • #3
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:
  • #4
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.
 
  • #5
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
 

1. What is torque and how is it different from force?

Torque is a measure of the rotational force or moment applied to an object. It is different from force in that force causes linear motion, while torque causes rotational motion.

2. How is torque calculated?

Torque is calculated by multiplying the force applied to an object by the distance from the pivot point to the point where the force is applied. The formula for torque is: Torque = Force x Distance.

3. What is static equilibrium and how does it relate to torque?

Static equilibrium refers to a state where an object is not moving or rotating. In order for an object to be in static equilibrium, the sum of all forces acting on the object must be equal to zero, and the sum of all torques acting on the object must also be equal to zero.

4. How do you solve a torque or static equilibrium problem?

To solve a torque or static equilibrium problem, you must first identify all the forces and torques acting on the object. Then, use the equations of equilibrium to set up and solve a system of equations to find the unknown forces or distances.

5. Can torque and static equilibrium be applied to real-world situations?

Yes, torque and static equilibrium are used in various fields of science and engineering to analyze and design structures, machines, and systems. They are also used in everyday situations, such as when opening a door or using a wrench to loosen a bolt.

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