Rigid Body Dynamics: Calculating Force on Beam Supports

In summary, in this conversation, a 3.0 m long rigid beam with a mass of 130 kg is discussed. It is supported at each end and a 65 kg student stands 2.0 m from the first support. The question is how much upward force each support exerts on the beam. After discussing different equations and methods, the final solution is found to be N1 = 849.333 N and N2 = 1061.667 N. The equations used include \Sigma F_x = 0, \Sigma F_y = 0, and net torque = 0.
  • #1
aligass2004
236
0

Homework Statement


A 3.0 m long rigid beam with a mass of 130 kg is supported at each end. A 65 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

http://i241.photobucket.com/albums/ff4/alg5045/p13-56.gif


Homework Equations





The Attempt at a Solution



I tried doing Fnet = N1 - W - N2 = 0, but that didn't work.
 
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  • #2
If you have to determine total upward force on the beam (N1+N2) than you have to equate net force on the beam = 0. In the equation written, you have missed out one of the weights out of the weight of the man and weight of the beam (unless W represents sum of the two weights). However, if you need N1 and N2 separately, then take moments of the forces about two supports and then solve for N1 and N2.
 
  • #3
I was using a similar example in the book, which said that Fnet = n1 - w - n2 = 0. Then it said that n2 = d1w/(d1 + d2). I tried 1m ((130+65)9.8)/1+2, but that was wrong.
 
  • #4
The normal forces act upward...

so N1 + N2 - w = 0

now take the torque about any point of your choosing... set the torque = 0... that gives a second equation. choosing the right point can make the math a little easier...

2 equations, 2 unknowns N1 and N2... you can solve for both.
 
  • #5
I still don't understand.
 
  • #6
aligass2004 said:
I still don't understand.

What is the torque about the left support in terms of the forces and distances...?
 
  • #7
T = rF, and the F is N1 = d2w/(d1+d2)?
 
  • #8
aligass2004 said:
T = rF, and the F is N1 = d2w/(d1+d2)?

How did you get the N1 = d2w/(d1+d2)? I don't know if that's right or not...

N1 exerts no torque about the left support... because the distance from the support is 0.

start with the basics.

Torque about the left support = w*2 - N2*3 (taking clockwise positive counterclockwise negative).

do you see how I got this?

set this equal to zero:

0 = w*2 - N2*3

you also have the equation

0 = N1 + N2 - w (sum of forces in the y-direction must be 0)

solve these 2 equations to get N1 and N2.
 
  • #9
I think I see how you got the equation. Is the weight (mass of beam + mass of student)g?
 
  • #10
aligass2004 said:
I think I see how you got the equation. Is the weight (mass of beam + mass of student)g?

oh... I forgot the beam... I was using w for the student...

torque = (rod's mass)*g*1.5 + (student's mass)*g*2 - N2*3

0 = (rod's mass)*g*1.5 + (student's mass)*g*2 - N2*3
 
  • #11
For N2 I got 1061.667. To find N1 I know I use 0 = N1 + N2 - w, but what is w? Is it the same as above?
 
  • #12
aligass2004 said:
For N2 I got 1061.667. To find N1 I know I use 0 = N1 + N2 - w, but what is w? Is it the same as above?

I messed that up too just like the moment equation... I forgot the weight of the student... but you can fix the equation... we need the sum of the forces in the vertical direction... take up as positive down as negative...

what equation do you get?
 
  • #13
The sum of the forces in the vertical direction would include the two normal forces, but I'm unsure about how to factor in the weight. I imagine torque is going to be involved.
 
  • #14
aligass2004 said:
The sum of the forces in the vertical direction would include the two normal forces, but I'm unsure about how to factor in the weight. I imagine torque is going to be involved.

No torque is only for the torque equation...

Just simply add up the forces taking up as positive, down as negative... no tricks.
 
  • #15
If there are no tricks, then F = N1 + N2 - (mass of the beam + mass of the student)g.
 
  • #16
aligass2004 said:
If there are no tricks, then F = N1 + N2 - (mass of the beam + mass of the student)g.

yes, exactly... this has to add to zero.

N1 + N2 - (mass of the beam + mass of the student)g = 0.

In all these statics (no moving parts) problems... you'll generally use 3 equations.

[tex]\Sigma F_x = 0[/tex] (sum of forces in the x-direction = 0)
[tex]\Sigma F_y = 0[/tex] (sum of forces in the y-direction = 0)

and

net torque about any point = 0. (you are free to choose any point).
 
  • #17
Awesome. See, I can do the forces part...just not the torque.
 
  • #18
aligass2004 said:
Awesome. See, I can do the forces part...just not the torque.

think force*distance, and clockwise/counterclockwise... you'll get used to it. Did you solve both N1 and N2?
 
  • #19
Yep, N2 = 1061.667 and N1 = 849.333
 
  • #20
aligass2004 said:
Yep, N2 = 1061.667 and N1 = 849.333

cool. that's what I get.
 

1. What is rigid body dynamics?

Rigid body dynamics is a branch of mechanics that deals with the study of movement and motion of solid objects, with a focus on the effects of forces on the object's motion.

2. How is force calculated on beam supports?

The force on beam supports can be calculated using the principles of statics, where the sum of all forces acting on a rigid body must be equal to zero. This means that the external forces acting on the beam, such as the weight of the object it is supporting, must be balanced by the internal forces within the beam itself.

3. What factors affect the force on beam supports?

The factors that affect the force on beam supports include the weight of the object being supported, the length and dimensions of the beam, and the angle at which the beam is supported. Other external forces, such as wind or vibrations, can also affect the force on the beam supports.

4. How is the stability of a beam support determined?

The stability of a beam support is determined by the amount and distribution of the external forces acting on the beam, as well as the strength and structural integrity of the beam itself. An unstable beam support can be identified by excessive deflection or deformation, and it may require additional support or reinforcement to prevent failure.

5. What are some common applications of rigid body dynamics in engineering?

Rigid body dynamics has many applications in engineering, including the design and analysis of structures such as bridges, buildings, and machines. It is also used in the development of vehicles, such as airplanes and cars, to ensure their stability and structural integrity. Additionally, rigid body dynamics is used in the study of biomechanics to analyze the movement and forces on the human body.

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