A 74 kg adult sits at one end of a 8.5 m long board. His 35 kg child sits on the other end.Where should the pivot be placed so that the board is balanced, ignoring the board's mass?
To determine where the pivot should be placed so that the board is balanced, we need to consider the concept of torque. Torque is the rotational force applied around a fixed point.
In this scenario, we have an adult sitting on one end of the board and a child sitting on the other end. The torque applied by each person is determined by their weight (mass × gravity) multiplied by the distance from the pivot point.
Let's say the pivot point is located a distance of x meters from the adult. Since the total torque must be balanced, we have the equation:
Torque exerted by the adult = Torque exerted by the child
(Mass of adult × gravity × distance from pivot) = (Mass of child × gravity × distance from pivot)
Plugging in the given values:
(74 kg × 9.8 m/s² × x) = (35 kg × 9.8 m/s² × (8.5 m - x))
Simplifying the equation:
74x = 35(8.5 - x)
74x = 297.5 - 35x
Combining like terms:
109x = 297.5
Dividing both sides by 109:
x ≈ 2.73 m
Therefore, the pivot should be placed approximately 2.73 meters from the adult, on the side where the child is sitting, in order to balance the board.
Just take the torque from where the adult sits. So it will be: 0 = (Force of the pivot)*(x) - (25)*(g)*(9). The force of the pivot can be calculated by taking the total force on the board. Which is (the force of the pivot) - (The adult's weight) - (The child's weight).
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