Question

a 3.0 m long rigid beam with a mass of 100 kg is supported at each end. an 80 kg student stands 2.0 m from support 1. how much upward force does each support exert on the beam?

Answer 1

`752.1\; {\rm N}` from support
`\texttt{1}` (
`2.0\; {\rm m}` from the student.)

`1013.7\; {\rm N}` from support
`\texttt{2}` (
`1.0\; {\rm m}` from the student.)

(Assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}`, the beam is level with negligible height, and that the density of the beam is uniform.)

Step-by-step explanation:

Weight of the beam:
`(100\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)}) = 981\; {\rm N}`.

Weight of the student:
`(80\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)}) = 784.8\; {\rm N}`.

Assuming that the beam is uniform. The center of mass of the beam will be
`(1/2)\, (3.0\; {\rm m}) = 1.5\; {\rm m}` away from each support.

Consider support
`\texttt{1}` as the fulcrum:

• For support
  `\texttt{2}` (with an upward force of
  `N_{\texttt{2}}`), the lever arm is
  `3.0\; {\rm m}`.
• For the center of mass of the beam (
  `981\; {\rm N}`), the lever arm is
  `1.5\; {\rm m}`.
• For the weight of the student (
  `784.8\; {\rm N}`), the lever arm is
  `2.0\; {\rm m}`.

Hence:

`\begin{aligned}N_{\texttt{2}}\, (3.0) = (981)\, (1.5) + (784.8) \, (2.0) \end{aligned}`.

`\begin{aligned}N_{\texttt{2}} &= ((981)\, (1.5) + (784.8) \, (2.0))/(3.0) \; {\rm N} = 1013.7\; {\rm N}\end{aligned}`.

In other words, support
`\texttt{2}` would exert an upward force of
`1013.7\; {\rm N}` on the beam.

Similarly, consider support
`\texttt{2}` as the fulcrum:

• For support
  `\texttt{1}` (with an upward force of
  `N_{\texttt{1}}`), the lever arm is
  `3.0\; {\rm m}`.
• For the center of mass of the beam (
  `981\; {\rm N}`), the lever arm is
  `1.5\; {\rm m}`.
• For the weight of the student (
  `784.8\; {\rm N}`), the lever arm is
  `(3.0 - 2.0)\; {\rm m} = 1.0\; {\rm m}`.

Hence:

`\begin{aligned}N_{\texttt{1}}\, (3.0) = (981)\, (1.5) + (784.8) \, (1.0) \end{aligned}`.

`\begin{aligned}N_{\texttt{1}} &= ((981)\, (1.5) + (784.8) \, (1.0))/(3.0) \; {\rm N} =752.1\; {\rm N}\end{aligned}`.

In other words, support
`\texttt{1}` would exert an upward force of
`752.1\; {\rm N}` on the beam.