A 10-ft,25lb seesaw is balanced by a little girl with a weight of 50lbs and her father with a weight of 200lbs at opposite ends. How far from the seesaw's center of mass must the fulcrum be placed?
50 (5 + d) + 25 d = 200 (5 - d)
250 + 75 d = 1000 - 200 d
275 d = 750
Well, it seems like we've got quite the balancing act going on here! Let's see, the little girl weighs 50lbs and her father weighs 200lbs. Now, assuming weight is directly proportional to mass, we can say that the little girl's weight-to-mass ratio is 2:1 and her father's is 8:1.
Now, the seesaw is balanced, so the total weight on each side of the fulcrum must be the same. Adding up the weight ratios on each side, we get:
2 (little girl) + 8 (father) = 10
This means that the fulcrum should be placed at a distance that divides the seesaw into two parts in a ratio of 2:8, or simplifying it, 1:4. Since the total length of the seesaw is 10 ft, the distance from the center of mass to the fulcrum should be:
10 ft ÷ (1 + 4) = 10 ft ÷ 5 = 2 ft
So, the fulcrum should be placed 2 feet from the center of mass. Happy balancing!
To balance the seesaw, the torques created by the weights on each side must be equal. Torque is calculated by multiplying the weight of an object by its distance from the fulcrum.
Let's assign variables to the unknowns:
- Distance of the fulcrum from the center of mass: x (in ft)
- Distance of the little girl from the center of mass: d1 = 10 ft
- Distance of the father from the center of mass: d2 = 0 ft (since the father is on the opposite side)
The torques can be calculated as follows:
1. Torque created by the little girl: T1 = 25 lb * d1
2. Torque created by the father: T2 = 200 lb * d2
Since the seesaw is balanced, T1 must equal T2:
25 lb * d1 = 200 lb * d2
Substituting the given values:
25 lb * 10 ft = 200 lb * d2
Solving for d2:
250 ft-lb = 200 lb * d2
d2 = 250 ft-lb / 200 lb
d2 = 1.25 ft
Therefore, the fulcrum must be placed 1.25 feet away from the center of mass to balance the seesaw.
To determine where the fulcrum should be placed, we need to consider the torque on each side of the seesaw.
Torque is the product of the force applied and the perpendicular distance from the fulcrum. In order to balance the seesaw, the torques on both sides should be equal.
Given:
- Little girl's weight = 50 lbs
- Father's weight = 200 lbs
- Seesaw length = 10 ft
Let's denote the distance of the fulcrum from the center of mass of the seesaw as 'x'.
The torque on the little girl's side is given by the product of her weight (50 lbs) and the distance from the fulcrum to her position (10 ft - x). So, her torque is 50 lbs * (10 ft - x).
Similarly, the torque on the father's side is given by the product of his weight (200 lbs) and the distance from the fulcrum to his position (x). So, his torque is 200 lbs * x.
Since the seesaw is balanced, the torque on both sides must be equal:
50 lbs * (10 ft - x) = 200 lbs * x.
To solve this equation, we can simplify it:
500 ft - 50x = 200x.
Rearranging the equation, we get:
250x = 500 ft.
Dividing both sides by 250, we find:
x = 2 ft.
Therefore, the fulcrum must be placed 2 feet from the center of mass of the seesaw.