An inelastic rope, with breaking tension of 250 N and length of 1 m, is tied between two walls at the same height. Calculate the greatest mass that can be hung from the middle of t... An inelastic rope, with breaking tension of 250 N and length of 1 m, is tied between two walls at the same height. Calculate the greatest mass that can be hung from the middle of the rope, without it breaking, if the walls are 0.6 m apart. A mass of 100 g is hanging from an inelastic string as shown in the diagram. Calculate the tension in each section, A and B, if the string is at angles of 30° and 60° respectively with the walls, as indicated.
Understand the Problem
The questions are asking to calculate forces related to an inelastic rope and an inelastic string. The first one asks for the maximum mass that can be suspended from the middle of the rope without breaking, given the tension and distance between the walls. The second question asks for the tension in each section of the string when a mass is hanging at specified angles.
Answer
The maximum mass is approximately $38.3 \, \text{kg}$. The tensions are $T_A \approx 0.58 \, \text{N}$ and $T_B \approx 1.15 \, \text{N}$.
Answer for screen readers
The maximum mass that can be hung from the middle of the rope without breaking is approximately 38.3 kg.
For question 13, the tensions are approximately:
$$ T_A \approx 0.58 , \text{N}, \quad T_B \approx 1.15 , \text{N} $$
Steps to Solve
- Identify Given Values for Question 12
The rope has a breaking tension of 250 N and a length of 1 m. The distance between the walls is 0.6 m.
- Determine the Weight Force
The maximum weight force ($F_g$) that can be applied to the rope can be calculated using the formula for tension. We know that:
$$ F_g = m \cdot g $$
where $g$ (acceleration due to gravity) is approximately $9.81 , \text{m/s}^2$.
- Set up the Equation for Tension
The tension in the rope when a mass is suspended at the midpoint creates two components. The vertical component of the tension must equal the weight force:
$$ T \sin(\theta) = F_g $$
To find $\theta$, we can use geometry. The rope forms a right triangle with half the distance between the walls (0.3 m) for the base and 0.5 m (half the length of the rope) as the hypotenuse.
- Calculate the Angle
Using the sine function:
$$ \sin(\theta) = \frac{0.3}{0.5} $$
Now, calculate $\theta$:
$$ \theta = \arcsin(0.6) $$
- Calculate the Maximum Mass
Insert the known values into the tension equation:
$$ F_g = T \sin(\theta) $$
Rearranging gives us:
$$ m = \frac{T \sin(\theta)}{g} $$
Inserting $T = 250 , \text{N}$ and $g = 9.81 , \text{m/s}^2$, we can find $m$.
- Identify Given Values for Question 13
We have a weight of 100 g (which is 0.1 kg) hanging at angles of 30° and 60°. Convert the mass to Newtons:
$$ F_g = m \cdot g = 0.1 , \text{kg} \cdot 9.81 , \text{m/s}^2 $$
- Set up Tension Components for Each Angle
For angle A (30°):
$$ F_g = T_A \sin(30°) + T_B \sin(60°) $$
Where the tension in each section is split using their respective angles.
- Solve for Tensions
Use the sine values:
$$ F_g = T_A \cdot 0.5 + T_B \cdot \frac{\sqrt{3}}{2} $$
This results in a system of equations.
- Substituting and Solving
You can solve for $T_A$ and $T_B$ using methods such as substitution or elimination, taking into account that $T_A \cos(30°) = T_B \cos(60°)$.
The maximum mass that can be hung from the middle of the rope without breaking is approximately 38.3 kg.
For question 13, the tensions are approximately:
$$ T_A \approx 0.58 , \text{N}, \quad T_B \approx 1.15 , \text{N} $$
More Information
The calculation shows how tension in a rope behaves when applied to a weight. The angles significantly change the distribution of tension, showing the importance of geometry in physical problems.
Tips
- Neglecting vertical components: Always remember to resolve forces into their components correctly.
- Incorrectly using angles: Make sure angles match the situation described, as wrong angles lead to incorrect tension calculations.
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