# A weight of 20 KN supported by two cords, one 3 m long and the other 4m long with points of support 5 m apart find the tensions T1 and T2 in the cords

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by venkateshsma

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by venkateshsma

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resolve t1 and t2 into horizontal and vertical components

t1 cos a+ t2 cos b=mg

and t1 sin a=t2 sin b

sin a=4/5 sin b= 3/5

4t1=3t2

take t1=3x and t2=4x

t1 cos a + t2 cos b=mg=20KN

3x (3/5)+4x(4/5)=20KN

5x=20

x=4

therefore t1= 12 KN t2 = 16 KN

sorry next time i will do the diagram in a nice way

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Let 3m cord make an angle t with horizontal and have tension T1. 3m, 4m, 5m long cords make a right angle triangle with 90 degrees at junction of T1 and T2. So T2 makes 180 - t - 90 = 90-t with horizontal.

T1 cos t = T2 cos 90-t => T1 = T2 sin t / cos t --- equation 1

In the triangle formed by two points of support and junction of T1 & T2, we have

3 sin t + 4 cos t = 5 (side joining two supports, its a right angle triangle)

divide by cos t => 3 tan t + 4 = 5 sec t

9 tan t * tan t + 16 + 24 tan t = 25 sec t * sec t = 24 ( 1 + tan t * tan t)

tan t = 3/4 by solving quadratic equation,

T1 = T2 * 3 /4.

T1 * T1 + T2 * T2 + 2 T1 T2 cos 90 = 20 * 20

T2 * T2 (3 * 3 / 4 * 4) + T2 * T2 = 400 => T2 * T2 = 400 * 16 / 25 = 256

T2 = 16 KN and T1 = 12 KN

T1 cos t = T2 cos 90-t => T1 = T2 sin t / cos t --- equation 1

In the triangle formed by two points of support and junction of T1 & T2, we have

3 sin t + 4 cos t = 5 (side joining two supports, its a right angle triangle)

divide by cos t => 3 tan t + 4 = 5 sec t

9 tan t * tan t + 16 + 24 tan t = 25 sec t * sec t = 24 ( 1 + tan t * tan t)

tan t = 3/4 by solving quadratic equation,

T1 = T2 * 3 /4.

T1 * T1 + T2 * T2 + 2 T1 T2 cos 90 = 20 * 20

T2 * T2 (3 * 3 / 4 * 4) + T2 * T2 = 400 => T2 * T2 = 400 * 16 / 25 = 256

T2 = 16 KN and T1 = 12 KN