<< Chapter < Page Chapter >> Page >

Discussion

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where F D size 12{F rSub { size 8{D} } } {} is less than 1/600th of the weight of the ship.

In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved.

Different tensions at different angles

Consider the traffic light (mass 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 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. 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. (d) Forces are shown with their components T sub one y and T sub two y pointing vertically upward. T sub one x points along the negative x direction, T sub two x points along the positive x direction, and weight W points vertically downward. (e) Vertical forces and horizontal forces are shown separately. Vertical forces T sub one y and T sub two y are shown by vector arrows acting along a vertical line pointing upward, and weight W is shown by a vector arrow acting downward. The net vertical force is zero, so T sub one y plus T sub two y is equal to W. On the other hand, 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. The net horizontal force is zero, so T sub one x is equal to T sub two x.
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 part (d) of the figure. There are two unknowns in this problem ( T 1 size 12{T rSub { size 8{1} } } {} and T 2 size 12{T rSub { size 8{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 size 12{F rSub { size 8{"net x"} } =T rSub { size 8{"2x"} } - T rSub { size 8{"1x"} } =0} {} .

Thus, as you might expect,

T 1 x = T 2 x size 12{T rSub { size 8{"1x"} } = T rSub { size 8{"2x"} } } {} .

This gives us the following relationship between T 1 size 12{T rSub { size 8{1} } } {} and T 2 size 12{T rSub { size 8{2} } } {} :

T 1 cos ( 30º ) = T 2 cos ( 45º ) size 12{T rSub { size 8{1} } "cos" \( "30"° \) =T rSub { size 8{2} } "cos" \( "45"° \) } {} .

Thus,

T 2 = ( 1 . 225 ) T 1 size 12{T rSub { size 8{2} } = \( 1 "." "225" \) T rSub { size 8{1} } } {} .

Note that T 1 size 12{T rSub { size 8{1} } } {} and T 2 size 12{T rSub { size 8{2} } } {} are not equal in this case, because the angles on either side are not equal. It is reasonable that T 2 size 12{T rSub { size 8{2} } } {} ends up being greater than T 1 size 12{T rSub { size 8{1} } } {} , because it is exerted more vertically than T 1 size 12{T rSub { size 8{1} } } {} .

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

F net y = T 1 y + T 2 y w = 0 size 12{F rSub { size 8{"net y"} } =T rSub { size 8{"1y"} } +T rSub { size 8{"2y"} } - w=0} {} .

This implies

T 1 y + T 2 y = w size 12{T rSub { size 8{"1y"} } +T rSub { size 8{"2y"} } =w} {} .

Substituting the expressions for the vertical components gives

T 1 sin ( 30º ) + T 2 sin ( 45º ) = w size 12{T rSub { size 8{1} } "sin" \( "30"° \) + T rSub { size 8{2} } "sin" \( "45"° \) =w} {} .

There are two unknowns in this equation, but substituting the expression for T 2 size 12{T rSub { size 8{2} } } {} in terms of T 1 size 12{T rSub { size 8{1} } } {} reduces this to one equation with one unknown:

T 1 ( 0 . 500 ) + ( 1 . 225 T 1 ) ( 0 . 707 ) = w = mg size 12{T rSub { size 8{1} } \( 0 "." "500" \) + \( 1 "." "225"T rSub { size 8{1} } \) \( 0 "." "707" \) =w= ital "mg"} {} ,

which yields

1 . 366 T 1 = ( 15 . 0 kg ) ( 9 . 80 m/s 2 ) size 12{ left (1 "." "366" right )T rSub { size 8{1} } = \( "15" "." "0 kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } {} .

Solving this last equation gives the magnitude of T 1 size 12{T rSub { size 8{1} } } {} to be

T 1 = 108 N size 12{T rSub { size 8{1} } ="108"" N"} {} .

Finally, the magnitude of T 2 size 12{T rSub { size 8{2} } } {} is determined using the relationship between them, T 2 size 12{T rSub { size 8{1} } } {} = 1.225 T 1 size 12{T rSub { size 8{2} } } {} , found above. Thus we obtain

T 2 = 132 N size 12{T rSub { size 8{2} } ="132 N"} {} .

Discussion

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).

Questions & Answers

what's acceleration
Joshua Reply
The change in position of an object with respect to time
Mfizi
how i don understand
Willam Reply
how do I access the Multiple Choice Questions? the button never works and the essay one doesn't either
Savannah Reply
How do you determine the magnitude of force
Peace Reply
mass × acceleration OR Work done ÷ distance
Seema
Which eye defect is corrected by a lens having different curvatures in two perpendicular directions?
Valentina Reply
acute astigmatism?
the difference between virtual work and virtual displacement
Noman Reply
How do you calculate uncertainties
Ancilla Reply
What is Elasticity
Salim Reply
using a micro-screw gauge,the thickness of a piece of a A4 white paper is measured to be 0.5+or-0.05 mm. If the length of the A4 paper is 26+or-0.2 cm, determine the volume of the A4 paper in: a). Cubic centimeters b). Cubic meters
Ancilla Reply
what is module
Alex Reply
why it is possible for an object(man) to stay on air without falling down?
akande Reply
its impossible, what do you mean exactly?
Ryan
Exactly
Emmanuella
it's impossible
Your
Why is it not possible to stand in air?
bikko
the air molecules are very light enough to oppose the gravitational pull of the earth on the man..... hence, freefall occurs
Arzail
what is physics
Joshua Reply
no life without physics ....that should tell you something
Exactly
Emmanuella
😎👍
E=MC^2
study of matter and energy and an inter-relation between them.
Minahil
that's how the mass and energy are related in stationery frame
Arzail
Ketucky tepung 10m
firdaus
Treeskin, 6m Cloud gam water 2m Cloud gam white 2m And buur
firdaus
Like dont have but have
firdaus
Two in one
firdaus
Okay
firdaus
DNA card
firdaus
hey am new over hear
Shiwani
War right? My impesilyty again. Don't have INSURAN for me
firdaus
PUSH
firdaus
I give
firdaus
0kay
firdaus
Hear from long
firdaus
Hehehe
firdaus
All physics... Hahahaha
firdaus
Tree skin and two cloud have tokside maybe
firdaus
Sold thing
firdaus
PUSH FIRST. HAHAHAAHA
firdaus
thanks
firdaus
Kinetic energy is the energy due to montion of waves,electrons,atoms, molecule,substances an object s.
Emmanuella
Opjective 0
firdaus
Atom nber 0
firdaus
SOME N
firdaus
10.000m permonth. U use momentom with me
firdaus
hi
Hilal
plz anyone can tell what is meteor and why meteor fall in night? can meteor fall in the day
Hilal
meteor are the glowy (i.e. heated when the enter into our atmosphere) parts of meteoroids. now, meteoroids are the debris resulting from the collision of asteroids or comets. yes, it occurs in daytime too, but due to the daylight, we cant observe it as clearly as in night
Arzail
thank's
Hilal
hello guys
Waka
wich method we use to find the potential on a grounded sphere
Noman
with out a physics the life is nothing to see
Yilma Reply
What do you want to talk about😋😋
Emmanuella
the study of all the natural events occuring around us..... this is Physics (until those events obey the laws of physics)
Arzail
Conservation of energy😰
Emmanuella
yeah, that too
Arzail
Energy, it always remains there in a physical system. it can only take the form either in motion (kinetic energy) or in rest (potential energy)
Arzail
In nature organisms feed on one another in an orderly way.
Emmanuella
that describes the food chain, in which we humans are at the top
Arzail
The energy that came initially from the sun 🌞is converted into a form in which it can be stored in green plant.
Emmanuella
Therefore, there is conservation of energy.
Emmanuella
DNA CARD
firdaus
"card"
firdaus
Darag
firdaus
What is x-ray
Daniel Reply
x-rays are electromagnetic Ray's produced when electrons with very high acceleration is brought to a stop by a target metal..
Felix
DNA CARD. DNA BLOOD(DARAH)
firdaus
@firdaus What is this DNA card? can I get to know?
Arzail
determine how much less the mass of lithium with mass number of 7 and proton of 3 nucleus is compared to that of its constituents.the mass of neutral Li 6.015123 u, calculate the total binding energy and the binding energy per nucleon
Barakat Reply
Try do car normally don't have oil. Like closing at all
firdaus
At
firdaus
Blosing design
firdaus
At-->automatic
firdaus
Blood DNA
firdaus

Get the best College physics course in your pocket!





Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask