A 5 m, 10 kg seesaw is balanced by a little girl(25 kg) and her father(80 kg) at the extreme opposite ends. How far from the seesaw's center of mass must the fulcrum be placed
what is the answer
To find the position of the fulcrum (or the distance from the center of mass), we can use the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Let's assume the distance of the fulcrum from the center of mass is represented by "x" meters.
The moment of an object is given by the product of its weight and the perpendicular distance from the pivot point (fulcrum).
The clockwise moments are given by the father's weight multiplied by his distance from the fulcrum: 80 kg * x meters = 80x kg·m.
The anticlockwise moments are given by the little girl's weight multiplied by her distance from the fulcrum: 25 kg * (5 - x) meters = 25(5 - x) kg·m.
Since the seesaw is balanced, these two moments are equal. Therefore, we can set up the equation:
80x = 25(5 - x).
Now, we solve the equation to find the value of x:
80x = 125 - 25x,
105x = 125,
x = 125 / 105,
x ≈ 1.19 meters.
Therefore, the fulcrum should be placed approximately 1.19 meters from the center of mass of the seesaw.
To determine the distance from the seesaw's center of mass where the fulcrum must be placed, we need to consider the principle of torque balance. Torque is the rotational equivalent of force, and for an object to be in rotational equilibrium, the torques on all sides of the fulcrum must be equal.
Let's denote the distance of the fulcrum from the center of mass as 'x'. The torque produced by the father is given by the product of his weight and the distance from his side to the fulcrum:
Torque by father = weight of father × distance from center of mass to fulcrum
= 80 kg × x
Similarly, the torque produced by the little girl is given by her weight multiplied by the distance from her side to the fulcrum:
Torque by little girl = weight of little girl × distance from center of mass to fulcrum
= 25 kg × (5 m - x)
Since the seesaw is balanced, the torques on either side of the fulcrum are equal. Therefore, we can equate the two torque equations:
80 kg × x = 25 kg × (5 m - x)
Simplifying the equation:
80x = 125 - 25x
Combining like terms:
105x = 125
Dividing both sides by 105:
x = 125 / 105
x ≈ 1.19 meters
Therefore, the fulcrum must be placed approximately 1.19 meters away from the center of mass of the seesaw.
It is rare to adjust the position of the fulcrum of a seasaw to adapt to persons of different masses.
Here we assume both Dad (D) and girl (G) are seated at the extremities of the 5m seasaw, and the centre of mass (M) of the 10 kg seasaw is in the middle. We assume the fulcrum (F) is between the Dad and the middle of the seasaw.
Draw a diagram,
G---------M----D
We know that GM=MD=2.5m
Denote MF=x
Take moments about the fulcrum,
Mg(2.5+x)+Ms(x)-Md(2.5-x)=0
where
Mg=25 kg (mass of girl)
Ms=10 kg (mass of seasaw)
Md=80 kg (mass of Dad)
Everything in the equation above is known except for x.
Solve for x.