Question

Two children are balanced on a seesaw that has a mass of 20.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. The center of mass of the seesaw is 0.153 m from the pivot (on the side of the first child)

(a) If the second child has a mass of 31.8 kg, how far (in m) is she from the pivot?
(b) What is the supporting force (in N) exerted by the pivot?
Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. (Submit a file with a maximum size of 1 MB.)

Answer 1

(a) d₂ = 1.404 m

(b) R = 762.44 N , Direction : vertical and downup

Step-by-step explanation:

Newton's first law

∑Mo=0

∑F=0

∑Mo : algebraic sum of moments around a point

∑F=0:algebraic sum of moments around a point

Formula to calculate M

Mo= F*d ( N*m)

F: Force (N)

d: Distance perpendicular to the force, from it to the pivot (m)

Data

m₁ =26.0 kg : mass of the first child

m₂ =31.8 kg kg : mass of the second child

m₃ = 20.0 kg : mass of the seesaw

W = m*g : weight (N)

m : mass (kg)

g : acceleration due to gravity(m/s²)

W₁ =26 kg*9.8m/s²= 254.8 N :weight of the first child

W₂ =31.8 kg*9.8m/s²=311.64 N : weight of the second child

W₃ = 20 kg*9.8m/s²= 196 N : weight of the seesaw

d₁ =1.60 m :Distance of the first child to the pivot

d₂ : Distance of the second child to the pivot

d₃ = 0.153 m : Distance of the seesaw to the pivot

Distance of the second child to the pivot (o)

∑Mo = 0

W₁*d₁+w₃*d₃-w₂*d₂= 0

(254.8)*(1.6) + (196*0.153) - (311.64)*d₂= 0

407.68+30 =(311.64)*d₂

437.68 = (311.64)*d₂

d₂ =437.68 /311.64

d₂ = 1.404 m

Static equilibrium of the forces

R: force (in N) exerted by the pivot

∑F=0

-W₁ - W₂-W₃ +R = 0

-254.8 -311.64-196 +R = 0

R = 762.44 N , Direction : vertical and downup