A uniform rod 30cm long is pivoted at it's center. A 40N weight is hung 5cm from the left end. where must the 50N weight be hung to maintain equilibrium?

Answer 1

To solve this problem, you would draw a diagram of the rod.

Then indicate the pivot and the first load.

Find the distance d1 between the first load P1 and the pivot: this is the lever arm for the first load.

To balance, then load P2 multiplied by d2 must equal P1*d1.

Since
d1=15-5=10 cm
P1=40 N
P2=50 N
and from
P1d1=P2d2
we solve to get
d2=P1d1/P2
=(40/50)*10
=8 cm. (from the pivot)

Answer 2

From where does that D1=15-5=10cm came

Answer 3

I think it's not a true answer😣😣

Answer 4

Half length of rod 15cm and weight is hung 5cm away FROM LEFT END so 15-5

Answer 5

D answer

Answer 6

Its correct he said hung from left end not pivot point so distance is 10cm 🙃

Answer 7

As the answer is 8 cm from the pivote than we have to find the distance at which 50N weight is hung than it is obtained by .subtracting 15-8=*7cm*
And i think it is correct distance when seen from right side where 50N weight is hung🙂

Answer 8

7cm from right end distance but 8cm distance from pivot point

Answer 9

But , Answer is 7 cm from the left end.