 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q9PE

Expert-verified Found in: Page 316 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A person carries a plank of wood $${\rm{2}}\;{\rm{m}}$$ long with one hand pushing down on it at one end with a force $${{\rm{F}}_{\rm{1}}}$$ and the other hand holding it up at $${\rm{50}}\;{\rm{cm}}$$ from the end of the plank with force $${{\rm{F}}_{\rm{2}}}$$. If the plank has a mass of $${\rm{20}}\;{\rm{kg}}$$ and its center of gravity is at the middle of the plank, what are the magnitudes of the forces $${{\rm{F}}_{\rm{1}}}$$ and $${{\rm{F}}_{\rm{2}}}$$?

The upward force is $${{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}$$.

The downward force is $${{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}$$.

See the step by step solution

## Step 1: Force

An external factor that can change the rest or motion of a body is the force.

## Step 2: Diagram

The mass is $${\rm{m = 20}}\;{\rm{kg}}$$.

The length of the plank is $${{\rm{r}}_{\rm{1}}}{\rm{ = 2}}{\rm{.00}}\;{\rm{m}}$$.

The distance between the forces is $${\rm{r = 0}}{\rm{.5}}\;{\rm{m}}$$. The free-body Diagram

## Step 3: Explanation

Zero linear acceleration and zero rotational acceleration are the first and second conditions, respectively.

Torque expression,

$${\rm{\tau = }}{{\rm{r}}_ \bot }{\rm{ \times F (1)}}$$

We begin by calculating the torque exerted by the plank's weight by putting our values for $${\rm{w = mg}}$$ and $${{\rm{r}}_{\rm{w}}}{\rm{ = l/2}}$$ into equation $${\rm{(1)}}$$, yielding:

$$\begin{array}{c}{{\rm{\tau }}_{\rm{w}}}{\rm{ = - }}\left( {\frac{{{\rm{2}}{\rm{.0\;m}}}}{{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{20\;kg \times 9}}{\rm{.8\;m/}}{{\rm{s}}^{\rm{2}}}} \right)\\{\rm{ = - 392\;N \times m}}\end{array}$$

$${{\rm{F}}_{\rm{1}}}$$ produces a torque of,

$$\begin{array}{c}{{\rm{\tau }}_{\rm{1}}}{\rm{ = }}{{\rm{r}}_{\rm{1}}}{\rm{ \times }}{{\rm{F}}_{\rm{1}}}\\{\rm{ = 0 \times }}{{\rm{F}}_{\rm{1}}}\\{\rm{ = 0}}\end{array}$$

And $${{\rm{F}}_{\rm{2}}}$$ produces a torque of,

$$\begin{array}{c}{{\rm{\tau }}_{\rm{2}}}{\rm{ = (0}}{\rm{.50\;m) \times }}{{\rm{F}}_{\rm{2}}}\\{\rm{ = 0}}{\rm{.50 }}{{\rm{F}}_{\rm{2}}}\end{array}$$

The second condition is satisfied since the system is in equilibrium. So, we get,

$$\begin{array}{c}\sum {\rm{\tau }} {\rm{ = }}{{\rm{\tau }}_{\rm{1}}}{\rm{ + }}{{\rm{\tau }}_{\rm{2}}}{\rm{ + }}{{\rm{\tau }}_{\rm{w}}}\\{\rm{ = 0}}\\{\rm{ - 392 + 0 + 0}}{\rm{.50}}{{\rm{F}}_{\rm{2}}}{\rm{ = 0}}\\{{\rm{F}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{392}}}}{{{\rm{0}}{\rm{.50}}}}\\{\rm{ = 784\;N}}\\{{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}\end{array}$$

The first condition has been met. So, we get,

$$\begin{array}{c}\sum {\rm{F}} {\rm{ = - }}{{\rm{F}}_{\rm{1}}}{\rm{ + }}{{\rm{F}}_{\rm{2}}}{\rm{ - mg}}\\{\rm{ = 0}}\\{{\rm{F}}_{\rm{1}}}{\rm{ = }}{{\rm{F}}_{\rm{2}}}{\rm{ - mg}}\\{\rm{ = 784 - (20 \times 9}}{\rm{.8)}}\\{\rm{ = 588\;N}}\\{{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}\end{array}$$

Hence, the forces are $${{\rm{F}}_{\rm{1}}}{\rm{ = 588\;N}}$$ and $${{\rm{F}}_{\rm{2}}}{\rm{ = 784\;N}}$$. ### Want to see more solutions like these? 