A 65-kg boy is sitting on a seesaw 0.6 m from the balance point. How far from the balance point should a 40-kg girl sit so that the seesaw remains balance?
The magnitude multiplied by the direction should be the same on both sides. So for the boy, it would be (65 kg x 0.6)= 39. Therefore the mass of the girl multiplied by the distance should equal 39 as well. So (39=X x 40). x = 0.975
To solve this problem, we need to understand that the seesaw will remain balanced if the torques on both sides of the pivot point are equal. In other words, the product of the weight of an object and its distance from the pivot (also called the lever arm) should be the same on both sides of the seesaw.
Let's assume that the 65-kg boy sits at a distance 'x' from the pivot point. We know that the weight (W) of an object is equal to its mass (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s². Therefore, the torque of the boy is given by:
Torque_boy = weight_boy * distance_boy_from_pivot
= (mass_boy * g) * x
= (65 kg * 9.8 m/s²) * x
= 637 N * x
Now, we need to find the distance at which the 40-kg girl should sit so that the seesaw remains balanced. Let's call this distance 'y'. The torque of the girl is given by:
Torque_girl = weight_girl * distance_girl_from_pivot
= (mass_girl * g) * y
= (40 kg * 9.8 m/s²) * y
= 392 N * y
For the seesaw to remain balanced, the torques on both sides should be equal. Therefore:
Torque_boy = Torque_girl
637 N * x = 392 N * y
Now, we can solve for 'y', which represents the distance the girl should sit from the balance point:
y = (637 N * x) / (392 N)
y = 1.626 x
So, the 40-kg girl should sit at a distance of 1.626 times the distance at which the 65-kg boy is sitting from the balance point in order to keep the seesaw balanced.