A uniform pole 7m long and weighing 10kg is supported by a boy 2m from one end and a man 3m from the other end.At what point must a 20kg weight be attached so that the man would support thrice as much weight as the boy?

Answer 1

from left end where x = 0
b up at x = 2.0
10 down at x = 3.5
m up at x = 4.0
20 down at x = x
and
m = 3 b
===================
forces up = forces down
b + m = 10 + 20 = 30
moments clockwise = moments counterclockwise about x = 0
10*3.5 + 20 x = b * 2 + m *4
so
20 x = 2 b + 4 m -35
b + m = 30
m = 3 b
-----------------------------
b + 3 b = 30
b = 30/4 = 15/2 = 7.5
m = 3 b = 22.5
then
20 x = 2 b + 4 m -35
20 x = 15 + 90 - 35 = 70
x = 7/2 = 3.5

LOL right at the middle
check
now we have 22.5+ 10 = 32.5 down at the middle
b*distance to middle = 22.5 * m's distance to middle ?
7.5 * (3.5-2) = 22.5 * (4.0 - 3.5)?
7.5 * 1.5 = 22.5 * .5 ?
YES ! Caramba :)
(should have seen it right from the start though)

Answer 2

thanks but is this answer correct

Answer 3

0.1

Answer 4

a uniform pole 7m long weighing 10kg is supported by a boy 2m from one end and a man 3m from the other end. at what point must a 20kg weight be attached so that man would support thrice as much weight as the boy

Answer 5

I want answer I’m confused here

Answer 6

How did u solve it

Answer 7

Am confused as well how come of 2.0, 4.0 , 3 .5 . Am not happy with this solving

Answer 8

This does not make muçh sense