A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A,and lies on horizontal ground. Calculate the force required to beging tto lift the end. Why would this force applied at the end A not be sufficient to lift the end A
A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A,and lies on horizontal ground. Calculate the force required to beging tto lift the end. Why would this force applied at the end A not be sufficient to lift the end A
To calculate the force required to begin lifting the end A of the pole, we need to consider the moment of the weight of the pole about point A.
The moment of a force is the force multiplied by the perpendicular distance from the point of rotation (in this case, point A). The moment is a measure of the turning effect of the force.
In this scenario, the weight of the pole (600N) acts downward at a distance of 4m from point A, causing a clockwise moment. To lift the end A, we need to apply a counterclockwise moment greater than the moment caused by the weight.
Now, let's calculate the moments:
Moment caused by the weight of the pole:
Moment(A) = Weight × Distance
Moment(A) = 600N × 4m
Moment(A) = 2400 Nm
To lift the end A, we need to apply a larger moment in the counterclockwise direction. If we apply a force F at the end A, the moment created will be:
Moment(Applied) = F × Distance(Applied)
Moment(Applied) = F × 10m
To lift the end A, the applied moment (F × 10m) must be greater than the moment caused by the weight (2400 Nm). Therefore, we can set up the equation:
F × 10m > 2400 Nm
To lift the pole, the force applied at end A must be greater than 240N (2400 Nm / 10m). If the force applied is less than 240N, the moment created will not overcome the moment caused by the weight, resulting in the end A not being lifted.