6.1 Solving problems with newton’s laws (Page 3/12)

University physics volume 1 Page 3 / 12

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 .

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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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