two children sit at opposite ends of a 60 lb. 10 ft. seesaw which is pivoted at its center one child weighs 50 lbs. and the other 75 lbs. the seesaw is uniform so that its center of gravity is at its geometric center, which also is its pivot point. A) the force exerted by the pivot point on the seesaw
To find the force exerted by the pivot point on the seesaw, we need to consider the principle of lever balance. In a balanced lever, the sum of the torques on either side of the pivot point is equal.
In this case, the torque is the product of the force and its perpendicular distance from the pivot point. Since the seesaw is in equilibrium (balanced), the sum of the torques on both sides of the pivot point must be zero.
Let's calculate the torques on each side of the seesaw:
Torque (clockwise) = Force(child1) * Distance(child1)
Torque (counterclockwise) = Force(child2) * Distance(child2)
In this scenario, the torque exerted by the child at one end of the seesaw cancels out the torque exerted by the child at the other end of the seesaw. This is because they are equidistant from the pivot point but have different weights.
Since the child's weight is a force acting vertically downward, we can use the formula: Torque = Force x Distance x sin(angle)
So, for both children, the torque will be: Torque = Weight x Distance(child) x sin(angle)
Now, let's do the calculations:
Child 1 (50 lbs.) torque: Torque(child1) = 50 lbs. x 5 ft. x sin(0°) = 0
Child 2 (75 lbs.) torque: Torque(child2) = 75 lbs. x 5 ft. x sin(0°) = 0
Since the torques on both sides of the seesaw cancel each other out, the total torque is zero. Therefore, the force exerted by the pivot point on the seesaw is also zero.
In summary, the force exerted by the pivot point on the seesaw is zero since the torques on both sides of the pivot point are equal and opposite, resulting in a balanced seesaw.