00:00Hi friends, your future starts with you.
00:04If you want to be successful, you have to learn.
00:09We see a particle connected by two ropes.
00:14Now, we are asked to determine the tension in one rope when the tension in the other rope is known.
00:23This animation might help you understand the problem.
00:28A particle is in the XY plane.
00:33This particle is connected by a rope to a pivot on the Z axis.
00:38The upper rope and the lower rope.
00:43Shortly after, the particle rotates around the axis.
00:48Of course, both ropes are under tension.
00:52We will calculate the tension in the ropes.
00:56This system is easier to see in a two-dimensional view.
01:03As usual, we will identify what forces are acting on the particle.
01:09A particle with mass around the Earth will experience the force of Earth's gravity downward.
01:17The ropes are not broken or slack, meaning there is a tension force on the rope.
01:24The particle is rotating horizontally, but there is no force acting in that direction.
01:31We must decompose the tension force in the rope into its component vectors.
01:37Theta here is the angle formed by the rope and the horizontal line.
01:45From the problem sheet, the length of the string is 13 meters.
01:50The distance from the particle to the axis is 12 meters.
01:55This length is 10 meters.
01:59Since this system is symmetric, we can divide this length into two equal parts.
02:06We see that the particle does not move vertically at all.
02:11The resultant force in the vertical direction is zero.
02:15T1 sine theta minus T2 sine theta equals mg.
02:21We can see the mass and tension of the upper string from the problem sheet.
02:28The value of sine theta itself can be seen in the figure.
02:35This factor is 120.
02:40This is a simple calculation.
02:44T2 is approximately 156 newtons.
02:49It turns out that the tension in the upper string is greater than the tension in the lower string.
02:57Happy learning, everyone!
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