Question

A woman who weighs 500 N stands on a 6.0-m-long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times the support force at the left end. How far from the right end is the woman standing?

Answer 1

Answer:

The woman's distance from the right end is 1.6m =

(8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.
Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.
Theses are the conditions for static equilibrium.

Answer 2

`1.2\; {\rm m}` from the right end of the board.

Step-by-step explanation:

The board can be considered as a lever supported on either the left end or the right end.

Let
`N_(L)` denote the support force on the left and. The support force at the right end would be
`N_(R) = 3\, N_(L)`.

Let
`x\; {\rm m}` be the distance between the person and the right end of the board. The distance from the left end of the board would be
`(6 - x)\; {\rm m}`.

Consider left end of the board as the fulcrum. Forces on the board include:

• Support force on the right end:
  `N_(R) = 3\, N_(L)` (upward.) Distance from fulcrum is
  `6\; {\rm m}`. Torque would be
  `6\, N_(R) = 18\, N_(L)` (counterclockwise.)
• Support force on the left end: torque would be
  `0` since this force is applied at a distance of
  `0\!` from the fulcrum.
• Weight of the board:
  `100\; {\rm N}` (downward.) Distance from fulcrum is
  `3\; {\rm m}`. Torque would be
  `(3\; {\rm m})\, (100\; {\rm N}) = 300\; {\rm N\cdot m}` (clockwise.)
• Weight of the person:
  `500\; {\rm N}` (downward.) Distance from the fulcrum (left end) is
  `(6 - x)\; {\rm m}`. Torque would be
  `(6 - x) \, (500)\; {\rm N\cdot m}` (clockwise.)

Similarly, consider right end of the board as the fulcrum. Forces on the board include:

• Support force on the right end: torque would be
  `0` since this force is applied at a distance of
  `0\!` from the fulcrum.
• Support force on the left end:
  `N_(L)` (upward.) Distance from fulcrum is
  `6\; {\rm m}`. Torque would be
  `6\, N_(R)` (clockwise.)
• Weight of the board:
  `100\; {\rm N}` (downward.) Distance from fulcrum is
  `3\; {\rm m}`. Torque would be
  `(3\; {\rm m})\, (100\; {\rm N}) = 300\; {\rm N\cdot m}` (counterclockwise.)
• Weight of the person:
  `500\; {\rm N}` (downward.) Distance from the fulcrum (left end) is
  `x\; {\rm m}`. Torque would be
  `(x) \, (500)\; {\rm N\cdot m}` (counterclockwise.)

In both situations, torque in the clockwise direction should balance torque in the counterclockwise direction.

With the left end of the board as the fulcrum:

`18\; N_(L) = 300 + (6 - x)\, (500)`.

With the right end of the board as the fulcrum:

`6\; N_(L) = 300 + (x)\, (500)`.

Solve this system of equations for the distance
`x`. Multiply both sides of the equation
`6\; N_(L) = 300 + (x)\, (500)` by
`3` to obtain:

`18\, N_(L) = 900 + (3\, x) \, (500)`.

Compare with the equation
`18\; N_(L) = 300 + (6 - x)\, (500)` to obtain:

`300 + (6 - x)\, (500) = 900 + (3\, x) \, (500)`.

`(6 - 4\, x)\, (500) = 600`.

`6 - 4\, x = (6 / 5)`.

`x = (24 / 20) = 1.2`.

In other words, the person should be standing at
`1.2\; {\rm m}` from the right end of the board.