A traffic sign is hanging from two wires. If the weight of the sign is 35 N, what are the tensions in the two wires as shown below? (# in degrees)
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\ 56 45/
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\ /
Reiny, I do not see how your calculation = 65.88, I get 25.2.
Also, I do not understand your diagram.
It looks like this
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\ 56 degree 45 degree /
\ /
\ /
\ /
\ /
\ /
\______________________/
| |
| 35 newtons |
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|_____________________|
As you can see, it is virtually impossible to draw a figure in this format.
I assume you can sketch a position diagram and a vector diagram of the problem.
My vector diagram has a vertical line of 35 N
the top angle is 34° (90-56)°
the bottom angle is 45°
and the angle opposite the 35 side is therefore 101°
using the sine law twice
string1/sin45 = 35/sin101
string1 = 35sin45/sin101 = 65.88
string2/sin34 = 35/sin101
string2 = 40.97
check my arithmetic
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\.56 degree ...........45 degree.../
\................................/
\............................../
\............................/
\........................../
\......................../
\______________________/
|.....................|
|........35 newtons...|
|.....................|
|_____________________|
To find the tensions in the two wires, we can use vector addition. The weight of the sign can be represented as a vector pointing straight down with a magnitude of 35 N.
First, we need to break down the weight vector into its horizontal and vertical components. The horizontal component will be 35 N * sin(56°), and the vertical component will be 35 N * cos(56°).
Next, we can analyze the forces in the horizontal direction. Since the wires are at an angle, the horizontal tensions in the wires will cancel out the horizontal component of the weight vector. Both wires will have the same tension, since they are symmetrically opposite.
Finally, we can analyze the forces in the vertical direction. The vertical tensions in the wires will support the weight of the sign, therefore, the sum of the vertical components of the tensions in the two wires must equal the vertical component of the weight vector.
To find the tension in each wire, we can set up the following equation:
2 * Tension * sin(45°) = 35 N * cos(56°)
Solving this equation will give us the tension in each wire.
Let's calculate it:
First, let's find the horizontal component of the weight vector:
Horizontal component = 35 N * sin(56°)
Horizontal component = 35 N * 0.829
Horizontal component ≈ 29.0 N
Now, let's calculate the vertical component of the weight vector:
Vertical component = 35 N * cos(56°)
Vertical component = 35 N * 0.559
Vertical component ≈ 19.6 N
Since the horizontal component of the weight vector is canceled out by the horizontal tensions in the wires, there is no horizontal tension in the wires.
Now, let's calculate the tension in each wire by rearranging the equation:
Tension * sin(45°) = (35 N * cos(56°)) / 2
Tension = ((35 N * cos(56°)) / 2) / sin(45°)
Substituting the values into the equation:
Tension ≈ ((35 N * 0.559) / 2) / 0.707
Tension ≈ 55.9 N / 0.707
Tension ≈ 79 N
Therefore, the tension in each wire is approximately 79 N.