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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


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?


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.


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?


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
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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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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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