Consider the figure below.
www.webassign.net/sercp9/4-p-017-alt.gif
(a) Find the tension in each cable supporting the 688-N cat burglar. (Assume the angle θ of the inclined cable is 38.0°.)
inclined cable - ???
horizontal cable - ???
vertical cable - ???
Well, ain't this a purr-fectly interesting problem! Let's unknot this together, shall we?
First, let's find the tension in the inclined cable. We can use a little trigonome-"meow"-try here. Given the angle θ of 38.0°, we can use the formula: tension = force / cosine(θ).
Since the force acting on the cat burglar is 688 N, we can plug in those numbers and calculate the tension in the inclined cable.
As for the horizontal cable, we're in for a "purr-suit" of the right answer. Since the horizontal cable is - you guessed it - horizontal, there's no vertical component to consider. Therefore, the tension in the horizontal cable is equal to the force acting on the cat burglar, which is 688 N.
Lastly, the tension in the vertical cable. As the name suggests, this cable is pulling "fur-reely" hard on the cat burglar, keeping them from falling. Since there's no horizontal component, the tension in the vertical cable is equal to the weight of the cat burglar, which is 688 N.
So, to sum it up:
Inclined cable: "Tension, you're on a slant!" - calculate using cosine(θ) and the given force.
Horizontal cable: "Tension, you're perfectly aligned!" - equal to the given force.
Vertical cable: "Tension, you're holding up the fort!" - equal to the weight of the cat burglar, which is 688 N.
I hope I've helped untangle this feline-filled problem for you!
To find the tension in each cable supporting the cat burglar, we can apply Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.
In this case, the cat burglar is not accelerating vertically, so the net force acting on them in the vertical direction is zero. Therefore, the sum of the vertical components of the tensions in the vertical and inclined cables must be equal to the weight of the cat burglar.
Given that the weight of the cat burglar is 688 N and the angle θ of the inclined cable is 38.0°, we can now calculate the tensions.
Let's label the tensions in the inclined, horizontal, and vertical cables as T_inclined, T_horizontal, and T_vertical, respectively.
Using trigonometry, we can find that the vertical component of the inclined cable's tension is T_inclined * cos(θ), and the horizontal component is T_inclined * sin(θ).
Since the vertical component of the inclined cable's tension and the tension in the vertical cable balance out the weight of the cat burglar, we have:
T_inclined * cos(θ) + T_vertical = 688 N
Similarly, the horizontal component of the inclined cable's tension and the tension in the horizontal cable must balance each other out, so:
T_inclined * sin(θ) + T_horizontal = 0
Now we have a system of two equations:
T_inclined * cos(θ) + T_vertical = 688 N (eq. 1)
T_inclined * sin(θ) + T_horizontal = 0 (eq. 2)
We can solve this system of equations to find the tensions.
From eq. 2, we have:
T_inclined * sin(θ) = - T_horizontal
T_inclined = - T_horizontal / sin(θ)
Now we can substitute this into eq. 1:
(- T_horizontal / sin(θ)) * cos(θ) + T_vertical = 688 N
- T_horizontal * cos(θ) + T_vertical * sin(θ) = 688 N * sin(θ)
Since sin(θ) and cos(θ) are known values (from the given angle θ = 38.0°), we can solve the equation for T_vertical:
T_vertical = (688 N * sin(θ) + T_horizontal * cos(θ)) / sin(θ)
Substituting this back into eq. 2, we get:
T_inclined * sin(θ) + T_horizontal = 0
(- T_horizontal / sin(θ)) * sin(θ) + T_horizontal = 0
- T_horizontal + T_horizontal = 0
0 = 0
This means that T_horizontal can have any value.
Therefore, the tension in the inclined cable (T_inclined) can be any value, and the tension in the vertical cable (T_vertical) is given by:
T_vertical = (688 N * sin(θ) + T_horizontal * cos(θ)) / sin(θ)
The tension in the horizontal cable (T_horizontal) can be any value, as long as it balances out the horizontal component of the tension in the inclined cable (T_inclined * sin(θ)).
So, we cannot determine the exact values of the tensions in each cable without knowing the tension in the horizontal cable (T_horizontal).
To find the tension in each cable supporting the cat burglar, you can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the sum of the forces exerted by the cables.
First, let's label the forces acting on the cat burglar. The inclined cable can be broken down into two components: the vertical component (T_v) and the horizontal component (T_h). The vertical cable exerts a force (T_v) directly upward, and the horizontal cable exerts a force (T_h) to the right.
To find the tension in each cable, we need to consider the forces in the vertical and horizontal directions separately.
In the vertical direction, the net force is equal to the weight of the cat burglar, which is given as 688 N. The forces acting in the vertical direction are the tension in the inclined cable (T_v) and the weight of the cat burglar (688 N). The weight acts downwards, so we can write the equation as:
T_v - 688 N = 0
Solving for T_v, we get:
T_v = 688 N
So the tension in the vertical cable is 688 N.
In the horizontal direction, the net force is equal to zero since the cat burglar is not accelerating horizontally. The force acting in the horizontal direction is the tension in the horizontal cable (T_h). So we can write the equation as:
T_h = 0
Therefore, the tension in the horizontal cable is zero (T_h = 0).
In summary, the tensions in each cable supporting the cat burglar are:
- Tension in the inclined cable (T_v): 688 N
- Tension in the horizontal cable (T_h): 0 N
- Tension in the vertical cable (T_v): 688 N
Please note that the figure you mentioned cannot be accessed by the AI. Consider providing a written description or providing additional details if needed.