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. Read more

Steps to Solve

1. Identify the situation and forces involved

The rope is held horizontally between two walls with a mass hanging at its center. The breaking tension of the rope is 250 N. The length of each half of the rope is 0.5 m and the distance between the walls is 0.6 m.

2. Determine the angle ( \theta )

To find the angle ( \theta ) formed by the rope at the points where it meets the mass, we can use trigonometry.

Using the opposite side (0.5 m) and the adjacent side (0.3 m, which is half of 0.6 m):

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

3. Calculate the angle ( \theta )

Use the inverse tangent function to get ( \theta ):

$$ \theta = \tan^{-1}\left(\frac{0.5}{0.3}\right) $$

4. Resolve the forces at the mass

The tension in the rope provides vertical support against the weight of the mass. Since there are two segments of rope, the vertical component ( T_y ) is:

$$ T_y = T \sin(\theta) $$

where ( T ) is the tension in one segment of the rope.

5. Set up the equation for the maximum tension

The maximum weight that the rope can hold corresponds to:

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

where:

• ( m ) is the mass.
• ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )).
• The maximum tension ( T ) is 250 N.

6. Rearranging for mass ( m )

Substitute ( T = 250 ) N into the equation, and solve for mass ( m ):

$$ m = \frac{2T \sin(\theta)}{g} $$

7. Calculate ( \sin(\theta) ) and find ( m )

First, calculate ( \theta ) using a calculator, then find ( \sin(\theta) ).

Finally, substitute the values into the mass equation to find ( m ).