a thick uniform rope of length 4m and mass 6kg is being pulled by 30N and 50N in opposite directions. what is the tension in the rope at point P which is 1.5m from 50N force?
Answers
We understand that tension is nothing but a force. we need to calculate the force at some point on the string.
One can consider the rope to be split into two parts of 1.5m and 2.5 mts and connected by a thin massless string. According to the question you need to find the tension in this string.
weight of the bigger part=6/4x2.5 Kg.
Acceleration of the whole part=20/6 m/s^2.
Think about this, if any body is moving, the acceleration on the whole will be same right? Conversely, if different points on any body are moving at different acceleration then they should be at different points after some time, meaning a brekdown of the body, hence not possible.
Applying this, the acceleration of the larger part will also be 20/6, one which we calculated above.
from this the net force on the larger body will be, F=ma, => (6/4x2.5)x(20/6)=12.5N
Now, this is the net force on the part of the rope. We also know that this part alreay has a force of 30N being applied to it initially, and the tension will be on the opposite direction as this force.
Thus net tension = 30+12.5=42.5N. (Option 3)
Sorry I could not attach a diagram of the same, please draw a figure (of the underlined text above) for better clarity.