A 130kg mail bag hangs by a vertical rope 3.9m long. A postal worker then displaces the bag to a position 2.2m sideways from its original position, always keeping the rope taut.
Well, well, well, looks like we have a postal worker putting some serious effort into that mail bag! Let's see what kind of circus act this is.
First things first, we need to find out how much tension is on that rope. It's like the heavyweight champ holding onto a rope for dear life.
Assuming the mail bag is in equilibrium, we can use the good ol' Pythagorean theorem to solve this riddle. The vertical distance and the horizontal distance form a right triangle, just like pizza and pineapple (controversial, I know).
Now, let's break out our math skills and get to solving! We'll find the tension in the rope, and by extension, how hard our postal worker is clowning around.
Using the Pythagorean theorem, we have:
(3.9m)^2 + (2.2m)^2 = (3.9m + 2.2m)^2
Simplifying:
15.21m^2 + 4.84m^2 = 6.1m^2
Combining like terms:
20.05m^2 = 6.1m^2
Uh-oh, it seems the numbers don't add up here. Either we've discovered the secrets of anti-gravity mail bags, or there might be some mistake in the equation. I'm afraid I can't provide a solid answer without a correction. My clown nose is gettin' all twisted up!
To solve this problem, we can use the concept of vectors in physics.
Step 1: Calculate the tension in the rope while the mail bag is hanging vertically.
The force of gravity acting on the mail bag can be calculated using the formula:
Weight = mass * acceleration due to gravity
Given that the mass of the mail bag is 130 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:
Weight = 130 kg * 9.8 m/s^2 = 1274 N
Since the mail bag is at rest, the tension in the rope must be equal to the weight of the bag. Therefore, the tension in the rope is 1274 N.
Step 2: Calculate the horizontal component of the tension when the mail bag is displaced.
When the mail bag is displaced horizontally, the vertical component of the tension remains the same. However, a new horizontal component is introduced.
Using trigonometric concepts, we can calculate the horizontal component of the tension.
Let's denote the horizontal component of the tension as T_horizontal.
T_horizontal = Tension * cos(theta)
To find theta, we can use the following equation:
tan(theta) = displacement / length of the rope
Given that the displacement is 2.2m and the length of the rope is 3.9m, we can calculate theta:
tan(theta) = 2.2m / 3.9m
theta = arctan(2.2m / 3.9m)
Using a calculator, we find that theta is approximately 31.6 degrees.
Now we can calculate the horizontal component of the tension:
T_horizontal = 1274 N * cos(31.6 degrees)
Using a calculator, we find that T_horizontal is approximately 1098 N.
Therefore, the horizontal component of the tension in the rope when the mail bag is displaced is approximately 1098 N.
To find the tension in the rope when the mailbag is displaced, we can break down the forces acting on it.
1. Gravitational Force: The weight of the mailbag acts vertically downward and is given by the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity (around 9.8 m/s^2).
Given that the mass of the mailbag is 130 kg, the weight is W = 130 kg * 9.8 m/s^2 = 1274 N (Newton).
2. Horizontal Components: When the mailbag is displaced sideways, the rope pulls back with a certain force to keep it taut. This force has two components:
- Horizontal Component: This component acts in the horizontal direction and is equal to the force required to displace the mailbag sideways. Let's call this force Fh.
- Vertical Component: This component acts in the vertical direction and counteracts a portion of the weight of the mailbag. It keeps the mailbag elevated at a constant height from the ground. Let's call this force Fv.
Since the rope is taut, the tension in the rope is the vector sum of these two components. Using Pythagoras' theorem, we can find the tension T in the rope:
T^2 = Fh^2 + Fv^2
Now we need to find Fh and Fv.
3. Finding Fh: The postal worker displaces the mailbag sideways by 2.2 m. We can calculate the horizontal force Fh using the formula:
Fh = m * a
Here, "m" is the mass of the mailbag and "a" is the acceleration due to gravity (9.8 m/s^2) since there is no other horizontal force acting on the mailbag.
Fh = 130 kg * 9.8 m/s^2 = 1274 N (Newton)
4. Finding Fv: To find Fv, we can use trigonometry. Since the rope is vertical, the vertical force Fv is equal to the vertical component of the tension in the rope. It can be given by:
Fv = T * cos(theta)
Here, theta is the angle between the rope and the vertical direction. In this case, since the bag is displaced horizontally, theta is the angle between the rope and the horizontal direction.
To find theta, we can use the Pythagorean theorem:
cos(theta) = adjacent / hypotenuse
In this case, the adjacent side is the horizontal displacement of the bag (2.2 m), and the hypotenuse is the length of the rope (3.9 m).
cos(theta) = 2.2 m / 3.9 m
theta = arccos(2.2 m / 3.9 m)
Using a scientific calculator or a trigonometric table, we can find the angle theta:
theta ≈ 52.85 degrees
Now we can calculate Fv using this angle:
Fv = T * cos(theta)
5. Finding Tension (T): Finally, we can substitute the values of Fh and Fv into the equation derived in step 2 to find the tension T.
T^2 = Fh^2 + Fv^2
T^2 = (1274 N)^2 + Fv^2
Now, we just need to substitute the values and calculate T.