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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

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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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