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?
I think it's not a true answer😣😣
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🙂
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)
From where does that D1=15-5=10cm came
D answer
7cm from right end distance but 8cm distance from pivot point
Its correct he said hung from left end not pivot point so distance is 10cm 🙃
Half length of rod 15cm and weight is hung 5cm away FROM LEFT END so 15-5
But , Answer is 7 cm from the left end.
To maintain equilibrium, the sum of the clockwise moments about the pivot point must equal the sum of the counterclockwise moments.
To solve this problem, we can use the principle of moments, which states that the moment of a force about a point is equal to the magnitude of the force multiplied by the perpendicular distance from the point to the line of action of the force.
Let's denote the distance between the pivot point and the left end of the rod as "x" and the distance between the pivot point and the 50N weight as "d".
The moment of the 40N weight about the pivot point is given by: 40N * 5cm
The moment of the 50N weight about the pivot point is given by: 50N * d
Since the rod is uniform and pivoted at its center, the length of the rod can be divided into two equal parts of 15cm each. Therefore, the distance between the pivot point and the center of the rod is 15cm.
The moment of the rod about the pivot point is given by: (15cm / 2) * 30cm
To maintain equilibrium, the sum of the clockwise moments must equal the sum of the counterclockwise moments:
40N * 5cm + 50N * d = (15cm / 2) * 30cm
To solve for "d", we can rearrange the equation:
50N * d = (15cm / 2) * 30cm - 40N * 5cm
50N * d = 225cm^2 - 200cm^2
50N * d = 25cm^2
d = (25cm^2) / 50N
d = 0.5cm
Therefore, the 50N weight must be hung 0.5cm from the left end of the rod in order to maintain equilibrium.