a framed painting weighs 220 newtons. the 2 wires support an angle of 120 degrees with each other. what are the tensions on the wire?
wires are 60 degrees from vertical
so
T cos 60 = 110
T (1/2) = 110
T = 220 N
Well, let's wire up some humor to solve this problem! Considering the tension is high in that frame, I'm going to deduce the wire tensions through a comedy routine.
First, let's name the wires. We'll call them "Tension" and "Laughter." Now, both wires are trying to support the painting, symbolizing career choices: one is more serious, and the other is more fun-loving.
To find the tensions, we'll start with Tension's performance. Tension is pulling at an angle of 120 degrees, which is quite off-center, much like a comedian telling a risky joke. When Tension takes the stage, it shares some of the load, let's say x newtons.
Next up is Laughter, pulling at the same angle. Laughter wants to keep things light and airy, so it also supports part of the painting, let's say y newtons.
Now, remember that the sum of all forces pulling upward must equal the weight of the painting, which is 220 newtons. That will be our punchline!
So, writing the equation for Tension's act: x + y = 220.
And considering their comedic relationship, using their shared angle of 120 degrees, we can apply a bit of trigonometry to define the relationship between x and y. The cosine of 120 degrees helps connect the two tensions.
Cosine(120) = (y / 220).
Now, let's solve the equation like a stand-up routine! Substituting the value of cosine(120) = -0.5, we've got:
-0.5 = (y / 220).
And multiplying both sides by 220 helps us find y, which is Laughter's tension:
y = -0.5 * 220 = -110.
But wait, tension is always positive! So the absolute value of y is:
|y| = 110 newtons.
Now, to determine Tension's tension, we'll just substitute this value back into the equation x + y = 220:
x + 110 = 220,
which simplifies to:
x = 220 - 110 = 110 newtons.
So, the tensions on the wire are 110 newtons for Tension and 110 newtons for Laughter. And together, they have a fantastic on-stage presence, keeping that painting hanging securely!
To find the tensions on the wires, we can break down the forces acting on the framed painting. Let's assume that T1 represents the tension in the first wire and T2 represents the tension in the second wire.
First, we need to resolve the weight of the painting into its components. Since the weight acts vertically downward, we can break it into two components: one along the direction of T1 and the other along the direction of T2.
The component of the weight along T1 can be found using trigonometry. We know that the angle between T1 and the weight is 120 degrees. Therefore, the component of the weight along T1 is given by:
Component of weight along T1 = weight × cos(angle between T1 and weight)
= 220 N × cos(120°)
Similarly, the component of the weight along T2 is given by:
Component of weight along T2 = weight × cos(angle between T2 and weight)
= 220 N × cos(120°)
Since the weight and tensions are in equilibrium, the sum of the vertical components of the tensions should equal the sum of the weight components perpendicular to the wires.
Sum of vertical components = Component of weight along T1 + Component of weight along T2
Now, let's calculate the component of the weight along T1 and T2:
Component of weight along T1 = 220 N × cos(120°)
= 220 N × (-0.5)
= -110 N
Component of weight along T2 = 220 N × cos(120°)
= 220 N × (-0.5)
= -110 N
Since the sum of the vertical components should equal zero, this means:
Sum of vertical components = T1 + T2 = 0
Substituting the values we calculated:
T1 + T2 = 0
T1 = -T2 (equation 1)
Next, let's consider the horizontal components of the tensions. The horizontal components will cancel each other out, as there is no horizontal acceleration in equilibrium.
So, the horizontal components of the tensions are zero:
Sum of horizontal components = 0
Since the angle between T1 and T2 is 120 degrees, the horizontal component of T1 and T2 will be equal:
Horizontal component of T1 = Horizontal component of T2
Let's denote the horizontal component of the tensions as H1 and H2:
H1 = H2 (equation 2)
From equation 1, we know T1 = -T2. From equation 2, we have H1 = H2.
Therefore, the tensions on the wires are as follows:
T1 = -T2
H1 = H2
To find the tensions on the wires, we need to use trigonometry and decompose the weight of the painting into its components along the wires. Let's call the tensions in the wires T1 and T2.
First, let's draw a free body diagram of the painting:
T1
/\
/ \
/ \
W /______\ T2
weight
We can see that the weight of the painting is acting vertically downwards. Now, let's decompose this weight into its components along the wires:
W
|\
| \
| \
| \
T1 | \ T2
| \
| \
|-------\
height
We know that the angle between the wires is 120 degrees. The component of weight acting along T1 can be found using cosine:
Component of weight along T1 = W * cos(60°)
Similarly, the component of weight acting along T2 can be found using cosine as well:
Component of weight along T2 = W * cos(60°)
Since the weight of the painting is given as 220 N, we can calculate the components of weight:
Component of weight along T1 = 220 N * cos(60°)
Component of weight along T2 = 220 N * cos(60°)
Now, to find the tensions, we need to set up the equilibrium equations for the vertical and horizontal directions:
For the vertical direction: T1 + T2 = W * sin(60°)
For the horizontal direction: T1 = T2
Since T1 = T2, we can substitute T1 for T2 in the vertical equation:
2T1 = W * sin(60°)
Now, let's solve for T1 (and T2):
2T1 = 220 N * sin(60°)
T1 = (220 N * sin(60°)) / 2
Similarly, T2 = (220 N * sin(60°)) / 2
Calculating these values should give you the tensions on the wires.