Chapter 15.2, Problem 1PP

Determine the impulse of the force for t = 2 s.

To determine

The impulse of the force for $t=2\text{s}$.

Answer to Problem 1PP

The impulse of the force for $t=2\text{s}$ is as follows:

1. a) $\underset{\_}{200\text{N}\cdot \text{s}}$.
2. b) $\underset{\_}{400\text{N}\cdot \text{s}}$
3. c) $\underset{\_}{12\text{N}\cdot \text{s}}$
4. d) $\underset{\_}{50\text{N}\cdot \text{s}}$
5. e) $\underset{\_}{160\text{N}\cdot \text{s}}$
6. f) $\underset{\_}{120\text{N}\cdot \text{s}}$

Explanation of Solution

Given:

The load on the block, $F=100\text{N}$.

Write the impulse of the force.

$I={\displaystyle {\int }_{t_1}^{t_2}Fdt}$ (I)

Here, change in time is dt, initial time is $t_1$, and final time is $t_2$.

(d)

Calculate the impulse of the force using the area under the F-t curve.

$\begin{array}{c}I=\text{Area}\\ =\left[\frac{1}{2}\left(t_1\times F\right)\right]+\left(t_2\times F\right)\end{array}$ (II)

Conclusion:

(a)

Apply the time limits from 0 to 2 s, substitute 100 N for F in Equation (I).

$\begin{array}{c}I={\displaystyle {\int }_0^{2\text{s}}\left(100\text{N}\right)dt}\\ =100{\left[t\right]}_0^{2\text{s}}\\ =100\left[2-0\right]\\ =200\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{\text{200}\text{N}\cdot \text{s}}$.

(b)

Apply the time limits from 0 to 2 s, substitute 200 N for F in Equation (I).

$\begin{array}{c}I={\displaystyle {\int }_0^{2\text{s}}\left(200\text{N}\right)dt}\\ =200{\left[t\right]}_0^{2\text{s}}\\ =200\left[2-0\right]\\ =400\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{\text{400}\text{N}\cdot \text{s}}$.

(c)

Apply the time limits from 0 to 2 s, substitute $\left(6t\right)\text{N}$ for F in Equation (I).

$\begin{array}{c}I={\displaystyle {\int }_0^{2\text{s}}\left(6t\text{N}\right)dt}\\ =6{\left[\frac{t^2}{2}\right]}_0^{2\text{s}}\\ =3\left[2^2-0\right]\\ =12\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{12\text{N}\cdot \text{s}}$.

(d)

Substitute 1 s for $t_1$, 20 N for $F$, and 3 s for $t_2$ in Equation (II).

$\begin{array}{c}I=\text{Area}\\ =\left[\frac{1}{2}\left(1\text{s}\times 20\text{N}\right)\right]+\left(3\text{s}\times 20\text{N}\right)\\ =50\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{50\text{N}\cdot \text{s}}$.

(e)

Apply the time limits from 0 to 2 s, substitute $80\text{N}$ for F in Equation (I).

$\begin{array}{c}I={\displaystyle {\int }_0^{2\text{s}}\left(80\text{N}\right)dt}\\ =80{\left[t\right]}_0^{2\text{s}}\\ =80\left[2-0\right]\\ =160\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{160\text{N}\cdot \text{s}}$.

(f)

Apply the time limits from 0 to 2 s, substitute $60\text{N}$ for F in Equation (I).

$\begin{array}{c}I={\displaystyle {\int }_0^{2\text{s}}\left(60\text{N}\right)dt}\\ =60{\left[t\right]}_0^{2\text{s}}\\ =60\left[2-0\right]\\ =120\text{N}\cdot \text{s}\end{array}$

Thus, the impulse of the force for $t=2\text{s}$ is $\underset{\_}{120\text{N}\cdot \text{s}}$.