A 5 m long see saw has a child whose mass is 25 kg sitting on one end. How far in meters from the center should a 70 kg adult sit on the opposite side for the see saw to balance?
Let A = the child and B = adult person.
To balance this beam, the moment of the right hand side (B) must equal the moment of the left hand side (A).
M(A) = M(B)
m(A)*g*2.5 = m(B)*g*d(B)
62.5 = 70*d(B)
=> d(B) = 70/62.5 = 1.12m
I hope this will help
To solve this problem, we need to consider the concept of torque. Torque is the rotational force that a force can create. In this case, we can calculate the torques on both ends of the seesaw to find the position where the seesaw will balance.
The torque on an object is calculated by multiplying the force applied to the object by the distance from the pivot point (or center of rotation). In this case, the distance from the pivot point is the distance of the person from the center of the seesaw.
Let's assume that the seesaw is balanced when the torques on both ends are equal.
Torque exerted by the child = Mass of the child × Acceleration due to gravity × Distance from the center
Torque exerted by the adult = Mass of the adult × Acceleration due to gravity × Distance from the center
Since the seesaw is balanced, the torques exerted by the child and the adult should be equal.
Mass of the child × Acceleration due to gravity × Distance from the center = Mass of the adult × Acceleration due to gravity × Distance from the center
Now we can cancel out the acceleration due to gravity and solve for the distance from the center (let's call it x).
Mass of the child × Distance of the child from the center = Mass of the adult × Distance of the adult from the center
25 kg × 5 m = 70 kg × x
Simplifying this equation, we get:
125 m = 70x
Now we can solve for x:
x = 125 m / 70
x = 1.79
So, the adult should sit approximately 1.79 meters from the center of the seesaw for it to balance.