A weight of 15 N is located at a distance of 2.5 meters from the fulcrum of a simple lever. At what distance from the fulcrum should a weight of 20.0 N be placed on the opposite side of the fulcrum to balance the system?
1.875 so about 1.9m
15N is applied at a distance of 2.5m
how many cm away should 20 N be placed? you only show M
To solve this problem, we can use the principle of moments in physics. The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.
In this case, the moments are the products of the weight and its distance from the fulcrum. So, we can set up the equation as follows:
Clockwise moment = Counterclockwise moment
For the given weight of 15 N at a distance of 2.5 meters:
Clockwise moment = 15 N * 2.5 m = 37.5 Nm
Now, let's assume the weight of 20 N is placed at a distance "x" from the fulcrum. The counterclockwise moment can be calculated as:
Counterclockwise moment = 20 N * x m = 20x Nm
To balance the system, the clockwise moment and counterclockwise moment must be equal:
37.5 Nm = 20x Nm
Now, we can solve for the "x" by dividing both sides of the equation by 20 N:
37.5 Nm / 20 N = x
x = 1.875 m
Therefore, the weight of 20 N should be placed at a distance of 1.875 meters from the fulcrum to balance the system.