In Fig. 6-45, a 1.34 kg ball is connected by means of two...

In Fig. 6-45, a $1.34 \mathrm{~kg}$ ball is connected by means of two massless strings, each of length $L=1.70 \mathrm{~m}$, to a vertical, rotating rod. The strings are tied to the rod with separation $d=1.70 \mathrm{~m}$ and are taut. The tension in the upper string is $35 \mathrm{~N}$. What are the (a) tension in the lower string, (b) magnitude of the net force $\vec{F}_{\text {net }}$ on the ball, and (c) speed of the ball? (d) What is the direction of $\vec{F}_{\text {net }} ?$

Transcript

00:02 In this question, in a figure, a 1 .35 kilogram of the ball, this is the ball, and this has a way of 1 .35 kilogram, ball is connected by the means of two massless string, each of length l, that is equal to 1 .7 meter.

00:18 So the length is equal to 1 .7 meter and the mass of the ball is equal to 1 .35 kilogram.

00:27 Kilogram now what we have to find in the first part we have to find the tension in the lower string and this road is rotating rod so the free body diagram of this question if i draw so it should be like let's say this is angle theta and this is my t u and this is the in this direction that will be t l this direction that will be t l and this direction and this direction and that will be t u this ball has a downward gravitational force that will be m g obviously right now how can i solve this question basically so the free body diagram for the ball is shown i already shown that is t u is the tension exerted by the upper string t u is the tension exited by the upper string and t l is the tension exerted by the lower string so we take the positive x direction to be leftward so basically according to the in the a part in the a part newton second law is t u cos of theta plus t l cos of theta minus m g should be equal to zero right and the second equation gives the tension in the lower string that will be since t l is equal to t u minus m g sign of theta now the question comes is what is theta in the question they have given that this triangle is equilateral triangle so i can easily say that theta will be equal to 30 degree right now we just have to put the values in this equation if i put the values so t l should be equal to 35 minus 1 .34 multiply by 9 .8 meter per second square divided by sign of theta that should be sign of 30 degree after solving this we will get the value of t l that is the tension lower string that will be approximately equal to 8 .74 newt this is the answer of the first part in the second part we have to find the net force in the second part the net force has a magnitude f net it should be equal to t u plus t l multiply by cost of theta multiply by multiply by 35 plus 8 .74 newton multiplied by cost of theta that is 30 degree...

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