In the figure above,A lever of length 200m is used to lift a load of mass 180kg. The pivot at P is 20m from the load. what minimum force F must be applied a...

In the figure above,A lever of length 200m is used to lift a load of mass 180kg. The pivot at P is 20m from the load. what minimum force F must be applied at the end of the lever [g = 10ms-2]

Answer Details

To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point. In this case, we can take moments about the pivot point P. Let's first calculate the weight of the load: Weight = mass x acceleration due to gravity Weight = 180 kg x 10 m/s^2 Weight = 1800 N Now, let's calculate the anticlockwise moment of the load about point P: Anticlockwise moment = weight x perpendicular distance from pivot to line of action of weight Anticlockwise moment = 1800 N x 20 m Anticlockwise moment = 36000 Nm To balance this moment, we need a clockwise moment produced by the force F applied at the end of the lever. The perpendicular distance from the pivot to the line of action of this force is 200 m - 20 m = 180 m. So, we can calculate the minimum force required as follows: Minimum force = clockwise moment / perpendicular distance from pivot to line of action of force Minimum force = 36000 Nm / 180 m Minimum force = 200 N Therefore, the minimum force required to lift the load is 200 N, which corresponds to option (D).