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. Read more

Steps to Solve

1. Determine the Forces Acting on the Rope

The maximum force that the rope can withstand before breaking is given as 250 N. This is the maximum tension, $T_{max}$.

2. Set Up the Geometry

The distance between the walls is 0.6 m, and the rope is 1 m long. Using the Pythagorean theorem, we can find the angle at which the rope is hanging. Let’s say the angle to each side is $\theta$.

Using the half-length of the rope for calculations:

$$ \sin(\theta) = \frac{0.3}{0.5} $$

3. Calculate Maximum Mass

The force due to gravity acting on the maximum mass, $m$, is expressed as $F = mg$, where $g$ is approximately $9.81 , \text{m/s}^2$. The maximum tension can be aligned with the component of the force acting through the angle.

The equation will be:

$$ T_{max} = 2T \sin(\theta) $$

4. Relate Mass and Tension

Setting the maximum tension equal to the gravitational force gives:

$$ mg = 2T \sin(\theta) $$

Substituting for $T$:

$$ m = \frac{T_{max} \cdot \sin(\theta)}{g} $$

5. Solve for Maximum Mass

Now, plugging in $T_{max} = 250 , \text{N}$ and calculating $\sin(\theta)$ from the geometric relation, we can find the maximum mass that can be supported.