High School

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A woman and her two children are playing on a seesaw. This seesaw has seats that can move to different distances from the fulcrum. Riders can also add seats to the seesaw. The woman weighs 145 lb, her son weighs 95 lb, her daughter weighs 70 lb, and each seat weighs 5 lb.

Question:
The woman is on the left side of the seesaw, 60 inches from the fulcrum. The daughter and son both get on the right side. The son sits 60 inches from the fulcrum. Where should the daughter sit to balance the seesaw? Please explain your process and give the correct answer.

Answer :

Answer:

40 inches

Step-by-step explanation:

Left side of seesaw:

Woman + sit = 145 lb + 5 lb = 150 lb

Distance from the fulcrum to the left side = 60 in

Moment of the force on the left side:

  • 150*60 = 9000

Right side of seesaw:

Son + sit = 95 lb + 5 lb = 100 lb

Distance from the fulcrum to the son's seat = 60 in

Moment of the force:

  • 100 * 60 = 6000

Daughter + sit = 70 lb + 5 lb = 75 lb

Distance from the fulcrum to the daughter's seat = x

Moment of the force:

  • 75 * x

Since left and right sides are balanced, moments should be equal on both sides.

This is shown as:

  • 75x+6000= 9000
  • 75x= 3000
  • x= 3000/75
  • x= 40

So, the daughter should sit at 40 in distance from the fulcrum.

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Rewritten by : Barada

Answer:

[tex]\large \boxed{\text{40 in from the fulcrum}}[/tex]

Step-by-step explanation:

An object is in rotational equilibrium when the net torque applied to it is zero.

For example, if a seesaw is not rotating, you know the torque on each side is balanced.

The counterclockwise torque of the woman equals the sum of the clockwise torques of her son and daughter.

The formula for torque τ is

τ = Fr,

where

F = the turning force and

r = the distance from the fulcrum

Thus, we can write

[tex]F_{\text{w}}r_{\text{w}} = F_{\text{s}}r_{\text{s}} + F_{\text{d}}r_{\text{d}}[/tex]

Data:

Each person has a 5 lb seat, so they are exerting 5 lb in addition to their weight.

Fw = 150 lb; rw = 60 in

Fₛ = 100 lb; rₛ = 60 in

Fd = 75 lb; rd = ?


Calculation:

[tex]\begin{array}{rcl}F_{\text{w}}r_{\text{w}} &=& F_{\text{s}}r_{\text{s}} + F_{\text{d}}r_{\text{d}}\\\text{150 lb $\times$ 60 in} & = & \text{100 lb $\times$ 60 in} + \text{75 lb} \times r_{\text{d}} \\\text{9000 in} & = & \text{6000 in}+ 75 r_{\text{d}}\\\text{3000 in} & = &75 r_{\text{d}}\\ r_{\text{d}} & = & \dfrac{\text{3000 in}}{75}\\\\& = & \textbf{40 in}\\\end{array}\\\text{The daughter should sit $\large \boxed{\textbf{40 in from the fulcrum}}$}[/tex]


Check:

[tex]\begin{array}{rcl}150 \times 60 &=& 100 \times 60 + 40 \times 75\\9000 & = & 6000 + 3000\\9000 & = & 9000\\\end{array}[/tex]

It checks.