Question

biomechanics problem

1. A see-saw on a playground has a 45kg child sitting 2.5m from its fulcrum. a. His dog has a mass of 30kg. How far does the dog need to sit away from the fulcrum on the opposite side of the see-saw to make it balance? b. The dog's three 5kg puppies join her on the see-saw. How far from the fulcrum does the group need to move for the see-saw to continue balancing?

Answer 1

Let

Mass of child (M) = 45kg

Length of child from fulcrum (L₁) = 2.5

Mass of dog (m) = 30 kg

Length of child from fulcrum (L₂)

As shown in the figure :

Child L2 Dog

We know that the see-saw will be balanced when torque about fulcrum or about point A is Zero.

Torque = Force × (perpendicular distance)

Part (a)

Child 12 Dpg VMg7 mg V

Net torque = MgL₁ (A.C.W) + mgL₂ (C.W) = 0

-MgL₁ (C.W) + mgL₂(C.W) = 0

ML L2 = - m 45 x 2.5 -= 3.75m 30

Part (b)

In this part it is not mentioned that, Does all the puppies has 5kg mass or each puppy has 5 kg mass

Lets check for both case :

Case 1:

Let mass of each puppy is 5kg

So, total mass of puppies = 3× 5 = 15 kg

Therefore, dog side mass will be (m^'​​​​​​) = 15 + 30 = 45 kg

As both side masses are equal so, the distance of group from fulcrum will be equal to distance of child from fulcrum

L₂ new = 2.5

So, group should be move towards fulcrum by distance :

D= L₂ - L₂ new

D = 3.75 - 2.5 = 1.25 m

Case 2:

If mass of all puppies is 5kg

then, total mass of dog side m' = 30 +5 = 35 kg

ML L2new = 45 x 2.5 m' 25 = 3.21m

So group should move towards fulcrum by distance :

D = L₂ - L_2new

D = 3.75 - 3.21

D = 0.54 m