A light weightless bar is provided at it centre and weight of 5n and 10n placed 3m and 2m, respectively from the pivot on one side are balanced by weight 20n weight on the other side how far in the 20n weight from the pivot.
5*3+10*2=20x
x= (15+20)/20=35/20=.... meters.
A light ( weightless) bar is pivoted at the center and weight of 5n and 10n placed 3m and 2m respectively from the pivot on one side and balanced by weight of 20n on the other side. How far is the 20n weight from the pivot
10*6 + 20*4 = 20x
60 + 80 = 20x
140 = 20x
x = 140/20
x = 7m
To solve this problem, we can use the principle of moments, also known as torque. Torque is the product of the force applied and the perpendicular distance from the pivot point.
In this case, we have a lightweight bar with a pivot point in the center. On one side of the pivot, we have weights of 5N and 10N placed 3m and 2m away from the pivot, respectively.
We need to find the distance from the pivot point to the 20N weight on the other side that will balance the system.
Let's assume that the distance from the pivot to the 20N weight is 'x' meters.
To establish equilibrium, the torque on one side should be equal to the torque on the other side:
Torque = Force × Distance
For the side with the 5N and 10N weights:
Torque1 = 5N × 3m + 10N × 2m
For the other side with the 20N weight:
Torque2 = 20N × x
Since the system is balanced, the torques on each side must be equal:
Torque1 = Torque2
5N × 3m + 10N × 2m = 20N × x
Simplifying the equation:
15N + 20N = 20N × x
35N = 20N × x
Now we can solve for x by dividing both sides of the equation by 20N:
x = 35N / 20N
x = 1.75 meters
Therefore, the 20N weight should be placed 1.75 meters away from the pivot point to balance the system.