This question aims to find the value of the smallest angle theta can make with a rope without breaking it by using laws of motion.
Consider a box of sweets weighing down the rope when people from across the buildings are sending this box. People from one building are sending this box of sweets to the people in the opposite building through a rope. When this box of sweets comes in the center of the rope, it makes an angle theta with the original position of the rope.
The position of this box of sweets in the center is not determined exactly. Both ends of the rope make an angle theta with the original position of the rope. We need to find the smallest angle among the two angles by applying Newton’s second law of motion.
Expert Answer
According to Newton’s second law of motion, any force acting on the body of mass m is equal to the rate of change of its velocity.
Applying Newton’s second law of motion:
\[ F = m a \]
Here, gravity is acting on the box of sweets so the acceleration will be equal to gravitational pull:
\[ F = m g \]
The force is acting along its vertical component so it will be written as:
\[ F _ y = 0 \]
\[ {\Sigma} F _ y = 0 \]
\[ 2 T sin \theta – m g = 0 \]
Tension in the rope is represented by T. It is a force acting on the rope when it is stretched.
\[ 2 T sin \theta = m g \]
To find an angle $ \theta $, we will rearrange the equation:
\[ sin \theta = \frac { m g } { 2 T } \]
Consider the mass of a box is 2 kg and it produces a tension of 30 N on the rope then the angle is:
\[ sin \theta = \frac { 2 \times 9 . 8 } { 2 \times 30 } \]
\[ sin \theta = \frac { 19 . 6 } { 60 } \]
\[ sin \theta = 0 . 3 2 6 \]
\[ \theta = sin ^ {-1} ( 0 . 3 2 6 ) \]
\[ \theta = 19 . 0 2 ° \]
Numerical Solution
The smallest angle acting on the rope without breaking it is 19.02°.
Example
Consider a person in the circus doing a stunt with the rope by hanging it. Both sides of this flexible rope are attached to the opposite cliffs. The mass of the person is 45 kg and the tension produced in the rope is 4200 N.
The smallest angle can be found by:
\[ {\Sigma} F _ y = 0 \]
\[ 2 T sin \theta – m g = 0 \]
Tension in the rope is represented by T. It is a force acting on the rope when it is stretched.
\[ 2 T sin \theta = m g \]
To find an angle $ \theta $, we will rearrange the equation:
\[ sin \theta = \frac { m g } { 2 T } \]
\[ sin \theta = \frac { 45 \times 9 . 8 } { 2 \times 4200 } \]
\[ sin \theta = \frac { 441 } { 8400 } \]
\[ sin \theta = 0 . 0 5 2 5 \]
\[ \theta = sin ^ {-1} ( 0 . 0 5 2 5 ) \]
\[ \theta = 3.00 ° \]
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