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Particle equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different tensions at different angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in [link] . Find the tension in each wire, neglecting the masses of the wires.

A sketch of a traffic light suspended from two wires supported by two poles is shown. (b) Some forces are shown in this system. Tension T sub one pulling the top of the left-hand pole is shown by the vector arrow along the left wire from the top of the pole, and an equal but opposite tension T sub one is shown by the arrow pointing up along the left-hand wire where it is attached to the light; the left-hand wire makes a thirty-degree angle with the horizontal. Tension T sub two is shown by a vector arrow pointing downward from the top of the right-hand pole along the right-hand wire, and an equal but opposite tension T sub two is shown by the arrow pointing up along the right-hand wire, which makes a forty-five degree angle with the horizontal. The traffic light is suspended at the lower end of the wires, and its weight W is shown by a vector arrow acting downward. (c) The traffic light is the system of interest, indicated by circling the traffic light. Tension T sub one starting from the traffic light is shown by an arrow along the wire making an angle of thirty degrees with the horizontal. Tension T sub two starting from the traffic light is shown by an arrow along the wire making an angle of forty-five degrees with the horizontal. The weight W is shown by a vector arrow pointing downward from the traffic light. A free-body diagram is shown with three forces acting on a point. Weight W acts downward; T sub one and T sub two act at an angle with the vertical. A coordinate system is shown, with positive x to the right and positive y upward. (d) Forces are shown with their components. T sub one is decomposed into T sub one y pointing vertically upward and T sub one x pointing along the negative x direction. The angle between T sub one and T sub one x is thirty degrees. T sub two is decomposed into T sub two y pointing vertically upward and T sub two x pointing along the positive x direction. The angle between T sub two and T sub two x is forty five degrees.  Weight W is shown by a vector arrow acting downward. (e) The net vertical force is zero, so the vector equation is T sub one y plus T sub two y equals W. T sub one y and T sub two y are shown on a free body diagram as equal length arrows pointing up. W is shown as a downward pointing arrow whose length is twice as long as each of the T sub one y and  T sub two y arrows. The net horizontal force is zero, so vector T sub one x is equal to minus vector T sub two x. T sub two x is shown by an arrow pointing toward the right, and T sub one x is shown by an arrow pointing toward the left.
A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical ( y ) and horizontal ( x ) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light.

Strategy

The system of interest is the traffic light, and its free-body diagram is shown in [link] (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in [link] (d). There are two unknowns in this problem ( T 1 and T 2 ), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

Solution

First consider the horizontal or x -axis:

F net x = T 2 x T 1 x = 0 .

Thus, as you might expect,

T 1 x = T 2 x .

This gives us the following relationship:

T 1 cos 30 ° = T 2 cos 45 ° .

Thus,

T 2 = 1.225 T 1 .

Note that T 1 and T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 ends up being greater than T 1 because it is exerted more vertically than T 1 .

Now consider the force components along the vertical or y -axis:

F net y = T 1 y + T 2 y w = 0 .

This implies

T 1 y + T 2 y = w .

Substituting the expressions for the vertical components gives

T 1 sin 30 ° + T 2 sin 45 ° = w .

There are two unknowns in this equation, but substituting the expression for T 2 in terms of T 1 reduces this to one equation with one unknown:

T 1 ( 0.500 ) + ( 1.225 T 1 ) ( 0.707 ) = w = m g ,

which yields

1.366 T 1 = ( 15.0 kg ) ( 9.80 m/s 2 ) .

Solving this last equation gives the magnitude of T 1 to be

T 1 = 108 N .

Finally, we find the magnitude of T 2 by using the relationship between them, T 2 = 1.225 T 1 , found above. Thus we obtain

T 2 = 132 N .

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
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please solve
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8m/s²
Aishat
What is Thermodynamics
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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
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50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
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definitely of physics
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what is field
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physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
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innocent
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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
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Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
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No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
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What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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