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Weights hanging on a beam

Using your graph board, draw a picture of a beam of length L hanging from a rope at its center. Hang a mass M1 from the left end of the beam. Hang a mass M2from the right end of the beam. Hang a mass M3 half way between M1 and the tie-point of the rope.

Assume that the system is in static equilibrium.

Part 1

Find the value of M3 given the following:

  • M1 = 1kg
  • M2 = 2kg

Solution:

Choose the tie-point of the rope as the axis of rotation.

The net torque about the axis of rotation is

Tnet = (L/2)*M1 + (L/4)*M3 - (L/2)*M2

For rotational equilibrium, the net torque must be equal to zero.

Rewriting and simplifying yields

L*(M1)/2 + L*(M3)/4 = L*(M2)/2

Dividing both sides by L yields

(M1)/2 + (M3)/4 = (M2)/2

Solving for M3 yields

M3 = 2*M2 - 2*M1

Substituting numeric values yields

M3 = 2*2kg - 2*1kg, or

M3 = 2kg

Part 2

If the mass of the beam is 2kg, what is the tension in the rope?

Solution:

In order for the system to be in translational equilibrium, the downward forces must be equal to the upward force exerted by the rope.

tension = (M1 + M2 + M3 + Mbeam)*g

Substituting numeric values yields

tension = (1kg + 2kg + 2kg + 2kg)*9.8m/s^2, or

tension = 68.6 newtons

Beam supported by a diagonal cable

Draw the following picture on your graph board. A vertical wall is on the left. A beam is attached to the wall with a hinge and extends outwardhorizontally to the right. (Without further support, the beam is free to rotate around the hinge causing the end of the beam to move up and down.)

To support the horizontal orientation of the beam, a cable is attached to the end of the beam and is attached to the wall above the attachment point of thebeam. The angle between the beam and the cable at the end of the beam is 30 degrees.

Assume that the system is in static equilibrium.

Assume that

  • the mass of the cable is negligible
  • the mass of the beam is given by M = 20kg
  • the length of the beam is L

Part 1

Draw vectors showing all of the forces that are exerted on the beam.

Solution:

The following forces are exerted on the beam:

  • A horizontal force pushing outward on the left end of the beam. Label it H.
  • A vertical force supporting the beam where it is connected to the wall by the hinge. Label it V.
  • A tension force in the cable. Label it T.
  • The weight of the beam acting downward at the center of the beam. Label it M*g.
  • The horizontal component of the tension force pushing the beam towards the wall. Label it T*cos(30 degrees).
  • The vertical component of the tension force pushing the end of the beam upward. Label it T*sin(30 degrees).

Part 2

What is the most judicious location for computing the sum of the torques to establish rotational equilibrium?

Solution:

This is probably a matter of opinion. However, if the torques are computed around the point where the beam attaches to the wall, three of the forces listedabove pass through that point and can be ignored when computing the sum of the torques around that point. Therefore, I consider that to be the mostjudicious point.

Part 3

Given the above assumptions , what are the magnitudes of the forces labeled V, H, and T required to produce translational androtational equilibrium?

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
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hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
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AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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