91Ó°ÊÓ

We need to find $F_{max}$which gives an impulse of $6.0Ns$.

A force is applied on the body for a short time, this force with time is called the impulse, so it is the area under the curve of the force-versus-time graph and it is the same as momentum. The impulse is the quantity $J_x$ and it is given by equation (11.6) in the form

$impulse=J_x={∫}_{t_i}^{t_f}{F}_x\left(t\right)dt$

=area under the $F_x\left(t\right)$curve between $t_i$and $t_{f_{}}$

A graph of force versus time is given in figure ex$11.5$where the force is given between the range of time from localid="1649327797637" $t_i=0ms$to $t_{f=}8ms$in the given figure, we can break the area into two triangles and one rectangle. The first triangle between $0ms$and $2ms$and height $F_{max}$so its area is

$$\begin{gathered} A_1=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}{F}_{max}\left(2×{10}^{-3}s\right) \\ ={10}^{-3}{F}_{max} \end{gathered}$$

The second triangle between $6ms$and $8ms$and height $F_{max}$so its area is

$$\begin{gathered} A_2=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}{F}_{max}\left(8×{10}^{-3}s-6×{10}^{-3}s\right) \\ ={10}^{-3}{F}_{max} \end{gathered}$$

The rectangle is between $2ms$and $6ms$with length localid="1649328673199" $F_{max}$so its area is

$$\begin{gathered} A_{rec\tan gle}=\left(length\right)\left(width\right) \\ =F_{max}\left(6×{10}^{-3}s-2×{10}^{-3}s\right) \\ =4×{10}^{-3}{F}_{max} \end{gathered}$$

Taking the summation of the three areas to get the impulse which is given by $J_x=6N.s$

$$\begin{gathered} J_x=A_{rec\tan gle}+A_1+A_2 \\ =4×{10}^{-3}{F}_{max}+{10}^{-3}{F}_{max}+F_{max} \\ =6×{10}^{-3}{F}_{max} \end{gathered}$$

Using the value of $J_x$and calculate the value of the m

ax force by

$$\begin{gathered} F_{max}=\frac{J_x}{6×{10}^{-3}} \\ =\frac{6N.s}{6×{10}^{-3}} \\ =1000N \end{gathered}$$