Math / Geometry

Question26. Find the tension in each of the three cables supporting the traffic light if it weighs $2.00 \times 10^{2} \mathrm{~N}$ .

Studdy Solution

STEP 1

1. The system is in static equilibrium, meaning the net force in all directions is zero.
2. The weight of the traffic light acts vertically downward.
3. The angles provided are with respect to the horizontal.
4. The tension in the vertical cable $T_3$ supports the entire weight of the traffic light.

STEP 2

1. Analyze the forces acting on the traffic light.
2. Set up equations based on equilibrium conditions.
3. Solve the equations to find the tensions in the cables.

STEP 3

Identify the forces acting on the traffic light. The forces are: - The weight of the traffic light, $W = 200 \, \mathrm{N}$ , acting downward. - The tension in cable $T_1$ , making an angle of $41.0^\circ$ with the horizontal. - The tension in cable $T_2$ , making an angle of $63.0^\circ$ with the horizontal. - The tension in cable $T_3$ , acting vertically upward.

STEP 4

Apply the equilibrium condition for vertical forces. The sum of the vertical components of the tensions must equal the weight of the traffic light:
$T_1 \sin(41.0^\circ) + T_2 \sin(63.0^\circ) + T_3 = 200 \, \mathrm{N}$

STEP 5

Apply the equilibrium condition for horizontal forces. The sum of the horizontal components of the tensions must be zero:
$T_1 \cos(41.0^\circ) = T_2 \cos(63.0^\circ)$

STEP 6

Since $T_3$ is vertical, it supports the entire weight of the traffic light. Therefore:
$T_3 = 200 \, \mathrm{N}$

STEP 7

Use the horizontal equilibrium equation to express $T_2$ in terms of $T_1$ :
$T_2 = \frac{T_1 \cos(41.0^\circ)}{\cos(63.0^\circ)}$

STEP 8

Substitute $T_2$ from STEP_5 into the vertical equilibrium equation:
$T_1 \sin(41.0^\circ) + \left(\frac{T_1 \cos(41.0^\circ)}{\cos(63.0^\circ)}\right) \sin(63.0^\circ) + 200 = 200$

STEP 9

Simplify and solve the equation from STEP_6 for $T_1$ :
$T_1 \left[\sin(41.0^\circ) + \frac{\cos(41.0^\circ) \sin(63.0^\circ)}{\cos(63.0^\circ)}\right] = 0$
$T_1 = 0$

STEP 10

Substitute $T_1 = 0$ back into the expression for $T_2$ :
$T_2 = \frac{0 \cdot \cos(41.0^\circ)}{\cos(63.0^\circ)} = 0$
The tensions in the cables are: - $T_1 = 0 \, \mathrm{N}$ - $T_2 = 0 \, \mathrm{N}$ - $T_3 = 200 \, \mathrm{N}$

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