Question

A 4 kg block is attached to a vertical rod by means of two strings of
equal length. When the system rotates about the axis of the rod, the
strings are extended as shown in figure.
(a) How many revolutions per minute must the system make in order
for the tension in the upper string to be 200 N?
(b) What is the tension in the lower string then?
CLOSE
Fig. 10.18​

Answer 1

Answer:

hope it will help you..

.....

Answer 2

Given:

Mass of the block = 4kg

Strings = 2

To Find:

(a) How many revolutions per minute must the system make for the tension in the upper string to be 200 N?

(b) What is the tension in the lower string then?

Solution:

Tension in lower wire  

Ti SinФ = T2sinФ + mg

= 200 x 4/5 = 4/5 x t2 + 40

= 160 = 4T2/5 + 40

T2 = 150N

Now,

SinФ = 4/5 and cosФ = 3/5

Horizontal direction = ma = T1 cos + T2 cos

Thus,

mrw² = ( T1 + T2) cos where OA= r = 3m

Hence,

4 x 3w² = ( 200 + 150 ) x 3/5

= 4.18

Employing 2πf = w

2πf = 4.18

f = 0.67 x 60

= 39.6

Answer: The tension is the lower string is 39.6 rpm